Number Theory, Ireland MO 2018, Problem 9

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Understand the problem

[/et_pb_text][et_pb_text _builder_version="4.0" text_font="Raleway||||||||" background_color="#f4f4f4" custom_margin="10px||10px" custom_padding="10px|20px|10px|20px" box_shadow_style="preset2"]The sequence of positive integers $a_1, a_2, a_3, ...$ satisfies $a_{n+1} = a^2_{n} + 2018$ for $n \ge 1$ .
Prove that there exists at most one $n$ for which $a_n$ is the cube of an integer.

Ireland MO 2018, Problem 9 [/et_pb_accordion_item][et_pb_accordion_item title="Topic" _builder_version="4.0" hover_enabled="0" open="off"]Number Theory [/et_pb_accordion_item][et_pb_accordion_item title="Difficulty Level" _builder_version="4.0" hover_enabled="0" open="off"]8/10 [/et_pb_accordion_item][et_pb_accordion_item title="Suggested Book" _builder_version="4.0" hover_enabled="0" open="off"]Excursion in Mathematics by Bhaskaryacharya Prathisthan [/et_pb_accordion_item][/et_pb_accordion][et_pb_text _builder_version="3.27.4" text_font="Raleway|300|||||||" text_text_color="#ffffff" header_font="Raleway|300|||||||" header_text_color="#e2e2e2" background_color="#0c71c3" custom_margin="48px||48px" custom_padding="20px|20px|20px|20px" border_radii="on|5px|5px|5px|5px" box_shadow_style="preset3"]

Start with hints

[/et_pb_text][et_pb_tabs active_tab_background_color="#0c71c3" inactive_tab_background_color="#000000" _builder_version="3.22.4" tab_text_color="#ffffff" tab_font="||||||||" background_color="#ffffff" hover_enabled="0"][et_pb_tab title="Hint 0" _builder_version="3.22.4"]Do you really need a hint? Try it first!

[/et_pb_tab][et_pb_tab title="Hint 1" _builder_version="4.0" hover_enabled="0"], wIt is so important to know and use the modulo technqiue at the right time. We will use the modulo technique, i.e. we will see the problem through the lens of modulo some number. What is that number? If you visit this website, you will understand that to handle cubes modulo something is 9. So, we will deal the whole equation modulo 9.

[/et_pb_tab][et_pb_tab title="Hint 2" _builder_version="4.0" hover_enabled="0"]

Definition: kth power residue of a number n is the complete residue system modulo n. For eg: Quadratic Residue (2nd power) of 4 is {0,1}.

Cubic(3rd) Power Residue of 9 is {0,1,-1}.
6th Power Residue of 9 is {0,1}
Quadratic(2nd Power) Residue of 9 is {0,1,4,7}

We will use these ideas here. [/et_pb_tab][et_pb_tab title="Hint 3" _builder_version="4.0" hover_enabled="0"]Let $a_k$ be the smallest integer which is a cube; let $a_k=a^3$ . Note that, $a_{k+1}=a^6+2018$ . Now, the modulo picture comes in. Starting from this cube. We will observe the sequence modulo 9. Case 1: $ a_k = 0$ mod 9 Then, the sequence modulo 9 will be $0 \mapsto 2 \mapsto 6 \mapsto 2 \mapsto \dots$ Hence, there are no further cubes possible as the cubic residues of 9 are {0,1,-1}. [/et_pb_tab][et_pb_tab title="Hint 4" _builder_version="4.0" hover_enabled="0"]Case 2: $ a_k = 1,-1$ mod 9 Then, the sequence modulo 9 will be $\pm 1 \mapsto 3 \mapsto 2 \mapsto 6 \mapsto 2 \mapsto \dots$ Hence, there are no further cubes possible as the cubic residues of 9 are {0,1,-1}. QED [/et_pb_tab][/et_pb_tabs][et_pb_text _builder_version="3.27.4" text_font="Raleway|300|||||||" text_text_color="#ffffff" header_font="Raleway|300|||||||" header_text_color="#e2e2e2" background_color="#0c71c3" custom_margin="48px||48px" custom_padding="20px|20px|20px|20px" border_radii="on|5px|5px|5px|5px" box_shadow_style="preset3"]

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Number Theory, France IMO TST 2012, Problem 3

Understand the problem

[/et_pb_text][et_pb_text _builder_version="4.0" text_font="Raleway||||||||" background_color="#f4f4f4" custom_margin="10px||10px" custom_padding="10px|20px|10px|20px" box_shadow_style="preset2"]Let $p$ be a prime number. Find all positive integers $a,b,c\ge 1$ such that:
$a^p+b^p=p^c.$ [/et_pb_text][/et_pb_column][/et_pb_row][et_pb_row _builder_version="3.25"][et_pb_column type="4_4" _builder_version="3.25" custom_padding="|||" custom_padding__hover="|||"][et_pb_accordion open_toggle_text_color="#0c71c3" _builder_version="3.22.4" toggle_font="||||||||" body_font="Raleway||||||||" text_orientation="center" custom_margin="10px||10px"][et_pb_accordion_item title="Source of the problem" _builder_version="4.0" open="on"]France IMO TST 2012, Problem 3[/et_pb_accordion_item][et_pb_accordion_item title="Topic" open="off" _builder_version="4.0"]Number Theory[/et_pb_accordion_item][et_pb_accordion_item title="Difficulty Level" _builder_version="4.0" open="off"]7/10 [/et_pb_accordion_item][et_pb_accordion_item title="Suggested Book" _builder_version="4.0" open="off"]Challenges and Thrills of Pre College Mathematics[/et_pb_accordion_item][/et_pb_accordion][et_pb_text _builder_version="3.27.4" text_font="Raleway|300|||||||" text_text_color="#ffffff" header_font="Raleway|300|||||||" header_text_color="#e2e2e2" background_color="#0c71c3" custom_margin="48px||48px" custom_padding="20px|20px|20px|20px" border_radii="on|5px|5px|5px|5px" box_shadow_style="preset3"]

Start with hints

[/et_pb_tab][et_pb_tab title="Hint 1" _builder_version="4.0" hover_enabled="0"]Observe that we will try to fundamental solutions to $ a^p + b^p = p^c$. A fundamental solution (a,b,c,p) gives infinitely many solutions $ (a.p^k, b.p^k, c+k, p)$. A fundamental solution is, therefore (a,b,c,p) if gcd(a,b) = 1. [/et_pb_tab][et_pb_tab title="Hint 2" _builder_version="4.0" hover_enabled="0"]We will focus on the fundamental solutions. We will deal with two cases: Case 1: p = 2 The equation reduces to $ a^2 + b^2 = p^2 $. As gcd(a,b) = 1, it implies a and b are odd. Now any odd square = 1 mod 4. So, $ a^2 + b^2 = 2 mod 4 $. Hence, the only fundamental solution is (1,1,1) = (a,b,c) We have that the following solutions are: $(1, 1, 1)$ and $\left(2^k, 2^k, 2^{2k+1}\right)$ .

[/et_pb_tab][et_pb_tab title="Hint 3" _builder_version="4.0" hover_enabled="0"]

Case 2: p is an odd prime This now requires the idea of Lifting the Exponents. Please read here if you don't know it. It is an advanced technique to deal with Diophantine Equations. Let's check that the conditions of the LTE are satisfying here. p is an odd prime. gcd(a,b) = 1. p doesn't divide a or b as we are looking for fundamental solutions. $ a^p = a mod p; b^p = b mod p $. Hence, $ a^p + b^p = a + b mod p $. So, p | a+b, and p don't divide a or b. Hence, we can apply LTE. [/et_pb_tab][et_pb_tab title="Hint 4" _builder_version="4.0" hover_enabled="0"]Using the LTE idea, we get $1+v_p(a+b)=v_p\left(a^p+b^p\right) = c$ Then since $a+b | a^p+b^p$ , we have that $a+b = p^{c-1}$ , so $a^p+b^p = pa+pb$ Now, you see this can't happen for large p, as the LHS is exploding too fast like exponential as p increases and RHS is linear in p. So, we will apply inequality to prove this and find a bound for p for which it works and search in that bound. Note that $a^p \geq pa, b^p \geq pb$ or $a^{p-1} \geq p, b^{p-1} \geq p$ if $a, b \geq 2$ . Therefore, we must have that $a=1, b=p^{c-1}-1$ (or vice versa). But clearly $\left(p^{c-1}-1\right)^p > p^c$ , so no solution. QED [/et_pb_tab][/et_pb_tabs][et_pb_text _builder_version="3.27.4" text_font="Raleway|300|||||||" text_text_color="#ffffff" header_font="Raleway|300|||||||" header_text_color="#e2e2e2" background_color="#0c71c3" custom_margin="48px||48px" custom_padding="20px|20px|20px|20px" border_radii="on|5px|5px|5px|5px" box_shadow_style="preset3"]

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Number Theory, South Africa 2019, Problem 6

Understand the problem

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Determine all pairs $(m, n)$ of non-negative integers that satisfy the equation
$20^m - 10m^2 + 1 = 19^n.$

[/et_pb_text][/et_pb_column][/et_pb_row][et_pb_row _builder_version="3.25"][et_pb_column type="4_4" _builder_version="3.25" custom_padding="|||" custom_padding__hover="|||"][et_pb_accordion open_toggle_text_color="#0c71c3" _builder_version="3.22.4" toggle_font="||||||||" body_font="Raleway||||||||" text_orientation="center" custom_margin="10px||10px"][et_pb_accordion_item title="Source of the problem" open="on" _builder_version="4.0"]South Africa MO 2019, Problem 6[/et_pb_accordion_item][et_pb_accordion_item title="Topic" _builder_version="4.0" open="off"]Number Theory[/et_pb_accordion_item][et_pb_accordion_item title="Difficulty Level" _builder_version="4.0" open="off"]7/10[/et_pb_accordion_item][et_pb_accordion_item title="Suggested Book" _builder_version="4.0" open="off"]Excursion in Mathematics by Bhaskaracharya Prathisthan[/et_pb_accordion_item][/et_pb_accordion][et_pb_text _builder_version="3.27.4" text_font="Raleway|300|||||||" text_text_color="#ffffff" header_font="Raleway|300|||||||" header_text_color="#e2e2e2" background_color="#0c71c3" custom_margin="48px||48px" custom_padding="20px|20px|20px|20px" border_radii="on|5px|5px|5px|5px" box_shadow_style="preset3"]

Start with hints

[/et_pb_text][et_pb_tabs active_tab_background_color="#0c71c3" inactive_tab_background_color="#000000" _builder_version="4.0" tab_text_color="#ffffff" tab_font="||||||||" background_color="#ffffff"][et_pb_tab title="Hint 0" _builder_version="3.22.4"]Do you really need a hint? Try it first!

[/et_pb_tab][et_pb_tab title="Hint 1" _builder_version="4.0"]If you read this, you will get to know some techniques to explore Diophantine Equations. Let's get an idea of m and n by using the modulo technique. To get an idea of n, we must remove or eliminate m, to do that we take modulo 10. Observe that the equation demands to be taken modulo 10, given the numbers and it turns out that $ 19^n = (-1)^n = 1 mod 10 $. It implies that n must be even. Try to get an idea of m now. Also, (0,0) is a solution. So, we take both m and n as non-zero.

[/et_pb_tab][et_pb_tab title="Hint 2" _builder_version="4.0"]

To remove n, using the information that n is even, we can remove the variable n, taking modulo 4. So, $ 2m^2 + 1 = 1 mod 4 $ implies m must be even. Let m = 2p.

Observe that RHS is a square and LHS $ < 20^{2p} $.

[/et_pb_tab][et_pb_tab title="Hint 3" _builder_version="4.0"]The largest square $ < 20^{2p}$ is $ (20^{p} - 1)^2$. Thus, $ LHS \leq (20^{p} - 1)^2 $. Hence $20^{2p} - 10(2p)^2 + 1 ~\le~ (20^p-1)^2 ~=~ 20^{2p}-2\cdot20^p+1$ ,
which simplifies to $p^2\ge20^{p-1}$ . (*) [/et_pb_tab][et_pb_tab title="Hint 4" _builder_version="4.0"]Now, this turns out to be inequality and this results in the solution p = 1. This gives the only solution (2,2) as (m,n).[/et_pb_tab][/et_pb_tabs][et_pb_text _builder_version="3.27.4" text_font="Raleway|300|||||||" text_text_color="#ffffff" header_font="Raleway|300|||||||" header_text_color="#e2e2e2" background_color="#0c71c3" custom_margin="48px||48px" custom_padding="20px|20px|20px|20px" border_radii="on|5px|5px|5px|5px" box_shadow_style="preset3"]

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Number Theory, Greece MO 2019, Problem 3

Understand the problem

[/et_pb_text][et_pb_text _builder_version="4.0" text_font="Raleway||||||||" background_color="#f4f4f4" custom_margin="10px||10px" custom_padding="10px|20px|10px|20px" box_shadow_style="preset2"]Find all positive rational $(x,y)$ that satisfy the equation : $yx^y=y+1$ [/et_pb_text][/et_pb_column][/et_pb_row][et_pb_row _builder_version="3.25"][et_pb_column type="4_4" _builder_version="3.25" custom_padding="|||" custom_padding__hover="|||"][et_pb_accordion open_toggle_text_color="#0c71c3" _builder_version="3.22.4" toggle_font="||||||||" body_font="Raleway||||||||" text_orientation="center" custom_margin="10px||10px"][et_pb_accordion_item title="Source of the problem" open="on" _builder_version="4.0"]Greece MO 2019, Problem 3[/et_pb_accordion_item][et_pb_accordion_item title="Topic" _builder_version="4.0" open="off"]Number Theory[/et_pb_accordion_item][et_pb_accordion_item title="Difficulty Level" _builder_version="4.0" open="off"]7/10[/et_pb_accordion_item][et_pb_accordion_item title="Suggested Book" _builder_version="4.0" open="off"]Challenges and Thrills of Pre College Mathematics[/et_pb_accordion_item][/et_pb_accordion][et_pb_text _builder_version="3.27.4" text_font="Raleway|300|||||||" text_text_color="#ffffff" header_font="Raleway|300|||||||" header_text_color="#e2e2e2" background_color="#0c71c3" custom_margin="48px||48px" custom_padding="20px|20px|20px|20px" border_radii="on|5px|5px|5px|5px" box_shadow_style="preset3"]

Start with hints

[/et_pb_text][et_pb_tabs active_tab_background_color="#0c71c3" inactive_tab_background_color="#000000" _builder_version="3.22.4" tab_text_color="#ffffff" tab_font="||||||||" background_color="#ffffff"][et_pb_tab title="Hint 0" _builder_version="3.22.4"]Do you really need a hint? Try it first!

[/et_pb_tab][et_pb_tab title="Hint 1" _builder_version="4.0"]Let us the given equation in terms of rational numbers and simplify the equation. Let $y=\frac{p}{q}$ and $x=\frac{r}{s}$ where $\gcd(p,q)=\gcd(r,s)=1$ . We have $p\cdot \left(\frac{r}{s}\right)^{\frac{p}{q}}=p+q \Longleftrightarrow p^q \cdot r^p=(p+q)^q \cdot s^p$ And since $\gcd(p+q,p)=\gcd(r,s)=1$ we must have $p^q=s^p$ and $r^p=(p+q)^q$ . Now, we have to find the solutions from these equations.[/et_pb_tab][et_pb_tab title="Hint 2" _builder_version="4.0"]Observe both the equations are of the form $x^a = y^b$. The idea is that due to the prime factorization theorem, we can specify that $x^a = y^b$ leads to a special form of the x and y. Claim: If $x^a=y^b$ for some $x,y,a,b$ naturals , then there exists a natural $z$ such that $x=z^m$ and $y=z^n$ where $m=\frac{b}{\gcd(a,b)}$ and $n=\frac{a}{\gcd(a,b)}$ .[/et_pb_tab][et_pb_tab title="Hint 3" _builder_version="4.0"]Proof of Claim: Consider the prime factorization theorem of the x and y in $x^a = y^b$. Observe that it implies x and y must have same set of primes by the prime factorization theorem. Let x and y contain the primes $p_1, p_2, ..., p_k$. Let $ x = \prod_{i=1}^{k} x_i$ and $ y = \prod_{i=1}^{k} y_i$. The above equation implies that $ a.x_i = b.y_i $. This implies that $ y_i = n.c $ and $ x_i = m.c $, where c is a natural number. Hence $ x = z^m, y = z^n$.

[/et_pb_tab][et_pb_tab title="Hint 4" _builder_version="4.0"]Using the intuitive claim, $p^q=s^p$ , there exists a $z$ such that $p=z^p$ , if $z \neq 1$ we have $p=z^p=z^{z^p}$ and continuing like this , $p$ is unbounded ,contradiction. So , we must have $z=1$ wich means $p=s=1$ and from $r^p=(p+q)^q$ we have $r=(q+1)^q$ So, the solutions come out to be $x=(q+1)^q$ and $y=\frac{1}{q}$ , where $q$ is any positive integer.[/et_pb_tab][/et_pb_tabs][et_pb_text _builder_version="3.27.4" text_font="Raleway|300|||||||" text_text_color="#ffffff" header_font="Raleway|300|||||||" header_text_color="#e2e2e2" background_color="#0c71c3" custom_margin="48px||48px" custom_padding="20px|20px|20px|20px" border_radii="on|5px|5px|5px|5px" box_shadow_style="preset3"]

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Algebra, Germany MO 2019, Problem 6

Understand the problem

Suppose that real numbers $x,y$ and $z$ satisfy the following equations: $\begin{align*} x+\frac{y}{z} &=2,\\ y+\frac{z}{x} &=2,\\ z+\frac{x}{y} &=2. \end{align*}$
Show that $s=x+y+z$ must be equal to $3$ or $7$ . Note: It is not required to show the existence of such numbers $x,y,z$ .

Germany MO 2019, Problem 6 [/et_pb_accordion_item][et_pb_accordion_item title="Topic" _builder_version="4.0" hover_enabled="0" open="off"]Algebra, Simultaneous Equations [/et_pb_accordion_item][et_pb_accordion_item title="Difficulty Level" _builder_version="4.0" hover_enabled="0" open="off"]6/10 [/et_pb_accordion_item][et_pb_accordion_item title="Suggested Book" _builder_version="4.0" hover_enabled="0" open="off"]Challenges and Thrills of Pre College Mathematics [/et_pb_accordion_item][/et_pb_accordion][et_pb_text _builder_version="3.27.4" text_font="Raleway|300|||||||" text_text_color="#ffffff" header_font="Raleway|300|||||||" header_text_color="#e2e2e2" background_color="#0c71c3" custom_margin="48px||48px" custom_padding="20px|20px|20px|20px" border_radii="on|5px|5px|5px|5px" box_shadow_style="preset3"]

Start with hints

[/et_pb_tab][et_pb_tab title="Hint 1" _builder_version="4.0" hover_enabled="0"]Observe that x = y = z = 1 gives a valid solution of the set of equations. In this case s = x+y+z = 3. Now, observe one thing that this set of equations is symmetric in (x,y,z). Observe that we are required to comment on (x+y+z). Rewriting the equations as: $xz+y = 2z, \qquad (1)$
$xy + z = 2x, \qquad (2)$
$yz + x = 2y \qquad (3)$
and then summing gives us that $x+y+z = xy + yz + zx = s.$ Our aim will be to reduce all the equations into a single variable. ( maybe a polynomial ). Let's consider the case, where all of x,y,z is not 1. [/et_pb_tab][et_pb_tab title="Hint 2" _builder_version="4.0" hover_enabled="0"]From now on we consider $x,y,z \neq 1$ . This also gives $x \neq y \neq z \neq x$ Solving the first expression $x=\frac{2z-y}{z}$ then plugging this into the second two gives:
$y+\frac{z^2}{2z-y}=2 \Rightarrow (2z-y)y+z^2=2(2z-y)$ $z+\frac{2z-y}{yz}=2 \Rightarrow yz^2+2z-y=2yz \Rightarrow y=-\frac{2z}{z^2-2z-1}$ as z is not equal to 1. Plugging the latter into the former and simplifying gives:
$\frac{z^2 (z^4-8z^3+14z^2-7)}{(z^2-2z-1)^2}=0 \Rightarrow z^4-8z^3+14z^2-7=0$ [/et_pb_tab][et_pb_tab title="Hint 3" _builder_version="4.0" hover_enabled="0"]Plugging the latter into the former and simplifying gives:
$\frac{z^2 (z^4-8z^3+14z^2-7)}{(z^2-2z-1)^2}=0 \Rightarrow z^4-8z^3+14z^2-7=0$ Now, observe that we already know z = 1 is a solution. This gives rise to $0=z^4-8z^3+14z^2-7=(z-1)(z^3-7z^2+7z+7) \Rightarrow z^3-7z^2+7z+7=0$ [/et_pb_tab][et_pb_tab title="Hint 4" _builder_version="4.0" hover_enabled="0"]Observe that the polynomial we have got in terms of z is also satisfied by x,y,z as the equations are symmetric in x,y,z. Hence we can claim that $ t^3 - 7t^2 + 7t + 7 = 0 $ has three solutions x,y,z. Hence, $ t^3 - 7t^2 + 7t + 7 = (t-x)(t-y)(t-z)$. Therefore, by Vieta's formula, x+y+z = 7. QED. [/et_pb_tab][/et_pb_tabs][et_pb_text _builder_version="3.27.4" text_font="Raleway|300|||||||" text_text_color="#ffffff" header_font="Raleway|300|||||||" header_text_color="#e2e2e2" background_color="#0c71c3" custom_margin="48px||48px" custom_padding="20px|20px|20px|20px" border_radii="on|5px|5px|5px|5px" box_shadow_style="preset3"]

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Combinatorics, Israel MO 2014, Problem 4

Understand the problem

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The three-digit number 999 has a special property: It is divisible by 27, and its digit sum is also divisible by 27. The four-digit number 5778 also has this property, as it is divisible by 27 and its digit sum is also divisible by 27. How many four-digit numbers have this property?

[/et_pb_text][/et_pb_column][/et_pb_row][et_pb_row _builder_version="3.25"][et_pb_column type="4_4" _builder_version="3.25" custom_padding="|||" custom_padding__hover="|||"][et_pb_accordion open_toggle_text_color="#0c71c3" _builder_version="3.22.4" toggle_font="||||||||" body_font="Raleway||||||||" text_orientation="center" custom_margin="10px||10px" hover_enabled="0"][et_pb_accordion_item title="Source of the problem" open="on" _builder_version="4.0" hover_enabled="0"]Israel 2014, Problem 4 [/et_pb_accordion_item][et_pb_accordion_item title="Topic" _builder_version="4.0" hover_enabled="0" open="off"]Combinatorics, Number Theory [/et_pb_accordion_item][et_pb_accordion_item title="Difficulty Level" _builder_version="4.0" hover_enabled="0" open="off"]6/10 [/et_pb_accordion_item][et_pb_accordion_item title="Suggested Book" _builder_version="4.0" hover_enabled="0" open="off"]Excursion in Mathematics by Bhaskarcharya Prathistan [/et_pb_accordion_item][/et_pb_accordion][et_pb_text _builder_version="3.27.4" text_font="Raleway|300|||||||" text_text_color="#ffffff" header_font="Raleway|300|||||||" header_text_color="#e2e2e2" background_color="#0c71c3" custom_margin="48px||48px" custom_padding="20px|20px|20px|20px" border_radii="on|5px|5px|5px|5px" box_shadow_style="preset3"]

Start with hints

[/et_pb_tab][et_pb_tab title="Hint 1" _builder_version="4.0" hover_enabled="0"]Let's write the problem mathematically, i.e. in terms of the equations. Let's write the condition mathematically. Let $abcd$ be a four-digit number, with $1\le a\le9$ and $0\le b,c,d\le 9$ , and $a,b,c,d$ positive integers. Then we need to have $1000a+100b+10c+d=27n$ and $a+b+c+d=27m$ , where $n,m$ are positive integers. Now, we have to count the number of such solutions. Observe that the sum of the digits can be at most 36. So, m = 1. Hence, a+b+c+d = 27. This leads to $111a+11b+c=3(n-1)$ . This implies $2b+c$ being a multiple of $3$ . Now, this implies we have quite a number of cases to investigate. Let's do them one by one patiently. [/et_pb_tab][et_pb_tab title="Hint 2" _builder_version="4.0" hover_enabled="0"]

b	c	a+d
0	3,6,9	27,24,21,18

Out of this b = 0, c = 9, a+d = 18, a = 9, c =9 is the only possibility considering the maximum possibble value of a+d being 18.

b	c	a+d
1	1,4,7	25,22,19

None of the cases is fine as the maximum possible value of a+d is 18. [/et_pb_tab][et_pb_tab title="Hint 3" _builder_version="4.0" hover_enabled="0"]

b	c	a+d
2	2,5,8	23,20,17

This gives rise to b = 2, c = 8, a+d = 17. Hence two solutions 8289 and 9288.

b	c	a+d
3	0,3,6,9	24,21,18,15

Thus you can understand this gives rise to 4 solutions: ( $6399,7398,8397$ and $9396$ ) [/et_pb_tab][et_pb_tab title="Hint 4" _builder_version="4.0" hover_enabled="0"]

b	c	a+d
4	1,4,7	22,19,16

This gives rise to three solutions: ( $7479,8478,$ and $9477$ ).

b	c	a+d
5	2,5,8	20,17,14

Only the last two choices are acceptable; the former gives us two solutions, and the latter 5 (for a total of seven solutions). [/et_pb_tab][et_pb_tab title="HInt 5" _builder_version="4.0" hover_enabled="0"]

b	c	a+d
6	0,3,6,9	21,18,15,12

The last three choices are acceptable and give us $1+4+7=12$ solutions.

b	c	a+d
7	1,4,7	19,16,13

The last two choices are acceptable and give us $3+6=9$ solutions.

b	c	a+d
8	2,5,8	17,14,11

All choices are acceptable and give us $2+5+8=15$ solutions.

b	c	a+d
9	0,3,6,9	18,15,15,9

All choices are acceptable and give us $1+4+7+9=21$ solutions. Therefore, there are $1+2+5+3+7+12+9+15+21=75$ four-digit numbers with this property. [/et_pb_tab][/et_pb_tabs][et_pb_text _builder_version="3.27.4" text_font="Raleway|300|||||||" text_text_color="#ffffff" header_font="Raleway|300|||||||" header_text_color="#e2e2e2" background_color="#0c71c3" custom_margin="48px||48px" custom_padding="20px|20px|20px|20px" border_radii="on|5px|5px|5px|5px" box_shadow_style="preset3"]

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Polynomial Functional Equation - Random Olympiad Problem

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Understand the problem

[/et_pb_text][et_pb_text _builder_version="3.27.4" text_font="Raleway||||||||" background_color="#f4f4f4" custom_margin="10px||10px" custom_padding="10px|20px|10px|20px" box_shadow_style="preset2" _i="1" _address="0.0.0.1"]Find all the real Polynomials P(x) such that it satisfies the functional equation: $latex P(2P(x)) = 2P(P(x)) + P(x)^{2} \forall real x $.

[/et_pb_text][/et_pb_column][/et_pb_row][et_pb_row _builder_version="3.25" _i="1" _address="0.1"][et_pb_column type="4_4" _builder_version="3.25" custom_padding="|||" custom_padding__hover="|||" _i="0" _address="0.1.0"][et_pb_accordion open_toggle_text_color="#0c71c3" _builder_version="3.22.4" toggle_font="||||||||" body_font="Raleway||||||||" text_orientation="center" custom_margin="10px||10px" _i="0" _address="0.1.0.0"][et_pb_accordion_item title="Source of the problem" open="on" _builder_version="3.29.2" _i="0" _address="0.1.0.0.0"]

Unknown [/et_pb_accordion_item][et_pb_accordion_item title="Topic" _builder_version="3.29.2" _i="1" _address="0.1.0.0.1" open="off"]Functional Equation, Polynomials[/et_pb_accordion_item][et_pb_accordion_item title="Difficulty Level" _builder_version="3.29.2" _i="2" _address="0.1.0.0.2" open="off"]7/10[/et_pb_accordion_item][et_pb_accordion_item title="Suggested Book" _builder_version="3.29.2" _i="3" _address="0.1.0.0.3" open="off"]Excursion in Mathematics Challenges and Thrills in Pre College Mathematics[/et_pb_accordion_item][/et_pb_accordion][et_pb_text _builder_version="3.27.4" text_font="Raleway|300|||||||" text_text_color="#ffffff" header_font="Raleway|300|||||||" header_text_color="#e2e2e2" background_color="#0c71c3" custom_margin="48px||48px" custom_padding="20px|20px|20px|20px" border_radii="on|5px|5px|5px|5px" box_shadow_style="preset3" _i="1" _address="0.1.0.1"]

Start with hints

[/et_pb_text][et_pb_tabs active_tab_background_color="#0c71c3" inactive_tab_background_color="#000000" _builder_version="3.29.2" tab_text_color="#ffffff" tab_font="||||||||" background_color="#ffffff" hover_enabled="0" _i="2" _address="0.1.0.2"][et_pb_tab title="Hint 0" _builder_version="3.22.4" _i="0" _address="0.1.0.2.0"]Do you really need a hint? Try it first!

[/et_pb_tab][et_pb_tab title="Hint 1" _builder_version="3.29.2" _i="1" _address="0.1.0.2.1"]Well, it is really good that the information polynomial is given! You should use that. What is the first thing that you check in a Polynomial Identity? Degree! Yes, check whether the degree of the Polynomial on both the LHS and RHS are the same or not. Yes, they are both the same $latex n^2 $. But did you observe something fishy? [/et_pb_tab][et_pb_tab title="Hint 2" _builder_version="3.29.2" _i="2" _address="0.1.0.2.2"]Now rewrite the equation as $latex P(2P(x)) - 2P(P(x)) = P(x)^{2}$. Do the Degree trick now... You see it right? Yes, on the left it is $latex n^2 $ and on the RHS it is $latex 2n $. So, there are two cases now... Figure them out!

[/et_pb_tab][et_pb_tab title="Hint 3" _builder_version="3.29.2" _i="3" _address="0.1.0.2.3"]

Case 1: $latex 2n = n^2 $... i.e. P(x) is either a quadratic or a constant function. Case 2: $latex P(2P(x)) - 2P(P(x)) $ has coefficient zero till $latex x^2n$. We will study case 1 now. Case 1: $latex 2n = n^2$... i.e. P(x) is either a quadratic or a constant function. $latex P(2P(x)) - 2P(P(x)) = P(x)^{2}$ = $latex P(2y) - 2P(y) = y^{2}$ where $latex y = P(x) $. Now, expand using $latex P(x) = ax^2 + bx +c$, it gives $latex 2ay^2 -c = y^2 $... Now find out all such polynomials satisfying this property. For e.g. $latex \frac{x^2}{2}$ is a solution. If P(x) is constant, prove that $latex P(x) = 0 / \frac{-1}{2} $. [/et_pb_tab][et_pb_tab title="Hint 4" _builder_version="3.29.2" _i="4" _address="0.1.0.2.4"]Case 2: $latex P(2y) - 2P(y) = y^{2}$. Assume a general form of P(x) = $latex $and show that P(x) must be quadratic or lesser degree by comparing coefficients as you have a quadratic on RHS and n degree polynomial of the LHS. Now, we have already solved it for quadratic or less degree. [/et_pb_tab][et_pb_tab title="Techniques Revisited" _builder_version="3.29.2" _i="5" _address="0.1.0.2.5"]

Always Compare the Degree of Polynomials in identities like this. It provides a lot of information.
Compare the coefficients of Polynomials on both sides to equalize the coefficient on both sides.

[/et_pb_tab][et_pb_tab title="Food for Thought" _builder_version="3.29.2" hover_enabled="0" _i="6" _address="0.1.0.2.6"]

Find all polynomials $latex P(2P(x)) - 8P(P(x)) = P(x)^{2}$.
Find all polynomials $ P(cP(x)) - d P(P(x)) = P(x)^{2} $ depending on the values of c and d.

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Connected Program at Cheenta

Math Olympiad is the greatest and most challenging academic contest for school students. Brilliant school students from over 100 countries participate in it every year. Cheenta works with small groups of gifted students through an intense training program. It is a deeply personalized journey toward intellectual prowess and technical sophistication.[/et_pb_blurb][et_pb_button button_url="https://cheenta.com/matholympiad/" url_new_window="on" button_text="Learn More" button_alignment="center" _builder_version="3.23.3" custom_button="on" button_bg_color="#0c71c3" button_border_color="#0c71c3" button_border_radius="0px" button_font="Raleway||||||||" button_icon="%%3%%" background_layout="dark" button_text_shadow_style="preset1" box_shadow_style="preset1" box_shadow_color="#0c71c3" _i="7" _address="0.1.0.7"][/et_pb_button][et_pb_text _builder_version="3.27.4" text_font="Raleway|300|||||||" text_text_color="#ffffff" header_font="Raleway|300|||||||" header_text_color="#e2e2e2" background_color="#0c71c3" custom_margin="50px||50px" custom_padding="20px|20px|20px|20px" border_radii="on|5px|5px|5px|5px" box_shadow_style="preset3" _i="8" _address="0.1.0.8"]

SMO(senior)-2014 Problem 2 Number Theory

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Understand the problem

[/et_pb_text][et_pb_text _builder_version="3.27.4" text_font="Raleway||||||||" background_color="#f4f4f4" box_shadow_style="preset2" custom_margin="10px||10px" custom_padding="10px|20px|10px|20px" _i="1" _address="0.0.0.1"]Find, with justification, all positive real numbers $a,b,c$ satisfying the system of equations: $a\sqrt{b}=a+c,b\sqrt{c}=b+a,c\sqrt{a}=c+b.$ [/et_pb_text][/et_pb_column][/et_pb_row][et_pb_row _builder_version="3.27.4" hover_enabled="0" _i="1" _address="0.1"][et_pb_column type="4_4" _builder_version="3.25" custom_padding="|||" custom_padding__hover="|||" _i="0" _address="0.1.0"][et_pb_accordion open_toggle_text_color="#0c71c3" _builder_version="3.27.4" toggle_font="||||||||" body_font="Raleway||||||||" text_orientation="center" custom_margin="10px||10px" hover_enabled="0" _i="0" _address="0.1.0.0"][et_pb_accordion_item title="Source of the problem" open="off" _builder_version="3.27" hover_enabled="0" _i="0" _address="0.1.0.0.0"]SMO (senior)-2014 stage 2 problem 2

[/et_pb_accordion_item][et_pb_accordion_item title="Topic" _builder_version="3.27" hover_enabled="0" _i="1" _address="0.1.0.0.1" open="on"]Number Theory[/et_pb_accordion_item][et_pb_accordion_item title="Difficulty Level" _builder_version="3.27.4" hover_enabled="0" _i="2" _address="0.1.0.0.2" open="off"]Easy [/et_pb_accordion_item][et_pb_accordion_item title="Suggested Book" _builder_version="3.27" hover_enabled="0" _i="3" _address="0.1.0.0.3" open="off"]Excursion in Mathematics

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Start with hints

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[/et_pb_tab][et_pb_tab title="Hint 1" _builder_version="3.27.4" hover_enabled="0" _i="1" _address="0.1.0.2.1"]Given all three relations are cyclic and symmetric . So without loss of generality it can be assumed that $ a \geq b \geq c >0 $ . Then proceed . [ Note $ (0, 0, 0) $ can't be a solution since $ a , b , c $ are positive reals .] [/et_pb_tab][et_pb_tab title="Hint 2" _builder_version="3.27.4" hover_enabled="0" _i="2" _address="0.1.0.2.2"]So $ a \sqrt b = a + c \Rightarrow a(\sqrt b - 1) = c \ [and \ we \ have \ a \geq c] \Rightarrow ( \sqrt b - 1 ) \leq 1 \Rightarrow b \leq 4 $[/et_pb_tab][et_pb_tab title="Hint 3" _builder_version="3.27.4" hover_enabled="0" _i="3" _address="0.1.0.2.3"]Similarly $ b \sqrt c = b + a \Rightarrow b(\sqrt c - 1) = a \ [and \ we \ have \ a \geq b ] \Rightarrow \sqrt c - 1 \geq 1 \Rightarrow c \geq 4 $[/et_pb_tab][et_pb_tab title="Hint 4" _builder_version="3.27.4" hover_enabled="0" _i="4" _address="0.1.0.2.4"]

Till now we have $ b \leq 4 \ and \ c \geq 4 $ , but we assumed that $ b \geq c $ . So it is clear that $ b = c =4 $ $ \Rightarrow a = 4 $ also. So the only triplet $ (a , b , c)$ is $ (4,4 ,4) $ . [/et_pb_tab][/et_pb_tabs][et_pb_text _builder_version="3.27.4" text_font="Raleway|300|||||||" text_text_color="#ffffff" header_font="Raleway|300|||||||" header_text_color="#e2e2e2" background_color="#0c71c3" border_radii="on|5px|5px|5px|5px" box_shadow_style="preset3" min_height="12px" custom_margin="50px||50px" custom_padding="20px|20px|20px|20px" _i="3" _address="0.1.0.3"]

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Math Olympiad is the greatest and most challenging academic contest for school students. Brilliant school students from over 100 countries participate in it every year. Cheenta works with small groups of gifted students through an intense training program. It is a deeply personalized journey toward intellectual prowess and technical sophistication.[/et_pb_blurb][et_pb_button button_url="https://cheenta.com/matholympiad/" url_new_window="on" button_text="Learn More" button_alignment="center" _builder_version="3.23.3" custom_button="on" button_bg_color="#0c71c3" button_border_color="#0c71c3" button_border_radius="0px" button_font="Raleway||||||||" button_icon="%%3%%" button_text_shadow_style="preset1" box_shadow_style="preset1" box_shadow_color="#0c71c3" background_layout="dark" _i="5" _address="0.1.0.5"][/et_pb_button][et_pb_text _builder_version="3.27.4" text_font="Raleway|300|||||||" text_text_color="#ffffff" header_font="Raleway|300|||||||" header_text_color="#e2e2e2" background_color="#0c71c3" border_radii="on|5px|5px|5px|5px" box_shadow_style="preset3" custom_margin="50px||50px" custom_padding="20px|20px|20px|20px" _i="6" _address="0.1.0.6"]

SMO (senior) -2014/problem-4 Number Theory

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Understand the problem

[/et_pb_text][et_pb_text _builder_version="3.27.4" text_font="Raleway||||||||" background_color="#f4f4f4" box_shadow_style="preset2" custom_margin="10px||10px" custom_padding="10px|20px|10px|20px" _i="1" _address="0.0.0.1"]For each positive integer $n$ let $x_n=p_1+\cdots+p_n$ where $p_1,\ldots,p_n$ are the first $n$ primes. Prove that for each positive integer $n$ , there is an integer $k_n$ such that $x_n<k_n^2<x_{n+1}$ [/et_pb_text][/et_pb_column][/et_pb_row][et_pb_row _builder_version="3.27" _i="1" _address="0.1"][et_pb_column type="4_4" _builder_version="3.25" custom_padding="|||" custom_padding__hover="|||" _i="0" _address="0.1.0"][et_pb_accordion open_toggle_text_color="#0c71c3" _builder_version="3.27" toggle_font="||||||||" body_font="Raleway||||||||" text_orientation="center" custom_margin="10px||10px" _i="0" _address="0.1.0.0"][et_pb_accordion_item title="Source of the problem" open="on" _builder_version="3.27" hover_enabled="0" _i="0" _address="0.1.0.0.0"]SMO (senior)-2014 stage 2 problem 4

[/et_pb_accordion_item][et_pb_accordion_item title="Topic" _builder_version="3.27" hover_enabled="0" _i="1" _address="0.1.0.0.1" open="off"]Number Theory[/et_pb_accordion_item][et_pb_accordion_item title="Difficulty Level" _builder_version="3.27" hover_enabled="0" _i="2" _address="0.1.0.0.2" open="off"]Medium[/et_pb_accordion_item][et_pb_accordion_item title="Suggested Book" _builder_version="3.27" hover_enabled="0" _i="3" _address="0.1.0.0.3" open="off"]Excursion in Mathematics

Start with hints

[/et_pb_tab][et_pb_tab title="Hint 1" _builder_version="3.27.4" hover_enabled="0" _i="1" _address="0.1.0.2.1"]We have $ P_1 = 2 , P_=3 , P_3=5 , P_4 =7 , P_5 = 11 \ and \ so \ on .... $. Now to understand the expression $x_n<k_n^2<x_{n+1}$ , observe . $ For \ n=1 \ , \ 2 < 2^2 < 2+3 $ $ For \ n=2 \ , \ 2+3 < 3^2 < 2+3+5 $ $ For \ n=3 \ , \ 2+3+5 < 4^2 < 2+3 +5+7$ $ For \ n=4 \ , \ 2+3 +5+7 < 5^2 <2+3 +5+7 +11 $ Now proceed to prove $ \forall n \geq 5 $ .[/et_pb_tab][et_pb_tab title="Hint 2" _builder_version="3.27.4" hover_enabled="0" _i="2" _address="0.1.0.2.2"]Observe $ \forall n \geq 5 $ we have $ P_n > (2n-1) $. [where $ n \in Z^+ $] Then try to use $ x_n = P_1 + P_2 + ...+P_5+..... +P_n > 1 +3 + ....+ 9 +... (2n-1) = n^2 \\ \Rightarrow x_n > n^2 , \forall n \geq 5[where \ n \in Z^+] $ .[/et_pb_tab][et_pb_tab title="Hint 3" _builder_version="3.27.4" hover_enabled="0" _i="3" _address="0.1.0.2.3"]Think if $ x_n=P_1 + P_2 + .... + P_5 +...P_n = b^2 for \ some \ n , b \in Z^+ $ , then we are done . If not so , then think $ m $ be the largest non negative integer such that $ (n+m)^2 < x_n $ . Now note that the next perfect square is $ (n+m+1)^2 $ . Observe that if we can prove that $ (n+m+1)^2 - (n+m)^2 = (2n+ 2m +1) \geq P_{n+1} $ , then we are done . Now try to verify this claim .[/et_pb_tab][et_pb_tab title="Hint 4" _builder_version="3.27.4" hover_enabled="0" _i="4" _address="0.1.0.2.4"]

Suppose our claim is not true i.e. $ P_n < 2n + 2m +1$ . So, $ P_n < 2n + 2m +1 \\ \Rightarrow 2n+ 2m \geq P_n , \forall n \in Z^+ \\ \Rightarrow (2n +2m-2)+(2n+ 2m -4)+.....2m \geq P_n + P_{n-1}+......+P_1 \\ \Rightarrow n^2 + 2mn -n \geq P_n + P_{n-1}+......+P_1 \\ \Rightarrow n^2 + 2mn -n \geq x_n \\ \Rightarrow n^2 + 2mn +m^2 > n^2 + 2mn -n\geq x_n \\ \Rightarrow (n+m)^2 > x_n $ . Contradiction! since we have assumed $ x_n = P_1 + P_2+......+P_{n-1} > (n+m)^2 $ . Thus ,$ (n+m+1)^2 \in (x_n , x_{n+1}) $ . [/et_pb_tab][/et_pb_tabs][et_pb_text _builder_version="3.27.4" text_font="Raleway|300|||||||" text_text_color="#ffffff" header_font="Raleway|300|||||||" header_text_color="#e2e2e2" background_color="#0c71c3" border_radii="on|5px|5px|5px|5px" box_shadow_style="preset3" min_height="12px" custom_margin="50px||50px" custom_padding="20px|20px|20px|20px" _i="3" _address="0.1.0.3"]

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Math Olympiad is the greatest and most challenging academic contest for school students. Brilliant school students from over 100 countries participate in it every year. Cheenta works with small groups of gifted students through an intense training program. It is a deeply personalized journey toward intellectual prowess and technical sophistication.[/et_pb_blurb][et_pb_button button_url="https://cheenta.com/matholympiad/" url_new_window="on" button_text="Learn More" button_alignment="center" _builder_version="3.23.3" custom_button="on" button_bg_color="#0c71c3" button_border_color="#0c71c3" button_border_radius="0px" button_font="Raleway||||||||" button_icon="%%3%%" button_text_shadow_style="preset1" box_shadow_style="preset1" box_shadow_color="#0c71c3" background_layout="dark" _i="5" _address="0.1.0.5"][/et_pb_button][et_pb_text _builder_version="3.27.4" text_font="Raleway|300|||||||" text_text_color="#ffffff" header_font="Raleway|300|||||||" header_text_color="#e2e2e2" background_color="#0c71c3" border_radii="on|5px|5px|5px|5px" box_shadow_style="preset3" custom_margin="50px||50px" custom_padding="20px|20px|20px|20px" _i="6" _address="0.1.0.6"]

Hidden triangular inequality (PRMO Problem 23, 2019)

Problem

Let ABCD be a convex cyclic quadrilateral . Suppose P is a point in the plane of the quadrilateral such that the sum of its distances from the vertices of ABCD is the least .If {PA,PB,PC,PD} = {3,4,6,8}.What is the maximum possible area of ABCD?

Topic

Geometry

Contest

Pre Regional Math Olympiad
2019

Reading
Mathematical circle .

Sequential Hints

Hint 1
Hint 2
Hint 3

The first step is very clear, you have to some how locate the point P .

claim : The intersection of the diagonal is the required point .
Proof : suppose that there is a point P' other than P (the intersection of diagonal) .
now try to apply triangular inequality in $\triangle$PBD & $\triangle$APC.
(do the rest yourself , and complete the proof)

so , the area of [ABCD]=$\frac{1}{2}$$\sin\theta$(PA*PB+PB*PC+PC*PD+PD*PA)
now, the area will maximum when $\sin\theta$=1
so , just put the value of PA, PB, PC,PD
You get the answer 55

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