Number Theory, Ireland MO 2018, Problem 9

[et_pb_section fb_built="1" _builder_version="3.22.4"][et_pb_row _builder_version="3.25"][et_pb_column type="4_4" _builder_version="3.25" custom_padding="|||" custom_padding__hover="|||"][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_padding="20px|20px|20px|20px" border_radii="on|5px|5px|5px|5px" box_shadow_style="preset3"]

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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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"][/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"]

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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Algebra, Austria MO 2016, Problem 4

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 $a,b,c\ge-1$ be real numbers with $a^3+b^3+c^3=1$ .
Prove that $a+b+c+a^2+b^2+c^2\le4$ , and determine the cases of equality.[/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"]Austria MO 2016. Final Round, Problem 4[/et_pb_accordion_item][et_pb_accordion_item title="Topic" _builder_version="4.0" open="off"]Inequality[/et_pb_accordion_item][et_pb_accordion_item title="Difficulty Level" _builder_version="3.22.4" open="off"][/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"]The idea is that you have to capture the symmetry in the equations and correspondingly find it. Observe that the inequality $a+b+c+a^2+b^2+c^2\le4$ with the constraint $a^3+b^3+c^3=1$ can be written as $ (a^3 - a ^2 - a +1) + (b^3 - b ^2 - b +1) + (c^3 - c ^2 - c +1) \ge 0 $ using the constraint. [/et_pb_tab][et_pb_tab title="Hint 2" _builder_version="4.0" hover_enabled="0"]Now, $ (a^3 - a ^2 - a +1) + (b^3 - b ^2 - b +1) + (c^3 - c ^2 - c +1) \ge 0 $ demands you to look into the polynomial $P(x)=(x+1)(x-1)^2=x^3-x^2-x+1$ . Thus, the problem reduces to show that if $a,b,c\ge-1$ are real numbers, then $P(a)+P(b)+P(c)\ge0$ . [/et_pb_tab][et_pb_tab title="Hint 3" _builder_version="4.0" hover_enabled="0"]What if we can show that individually if $x\ge-1$ we always have $P(x)\ge0$ ? Then our problem will be solved right? We have $P(x)=(x+1)(x-1)^2=x^3-x^2-x+1$ , observe that it automatically implies that if $ x+1 \geq 0 $ then we will have $ P(x) \geq 0$. [/et_pb_tab][et_pb_tab title="Hint 4" _builder_version="4.0" hover_enabled="0"]Equality Cases: For equality we must have $P(a)=P(b)=P(c)=0$ , and hence $a,b,c\in\{-1,+1\}$ .
Hence equality holds if and only if one of the three variables is $-1$ and the other two are $+1$ . 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, Cyprus IMO TST 2018, Problem 1

Understand the problem

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Determine all integers $n \geq 2$ for which the number $11111$ in base $n$ is a perfect square.

[/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"]Cyprus IMO TST 2018, Problem 1 [/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"]7/10 [/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"]Let us write the problem in Mathematical Language i.e. in the form of equations. $ (11111)_n$ in base n $ = 1 + n + n^2 + n^3 + n^4 $. So, the problem reduces to finding positive integer solutions to $ m^2 = 1 + n + n^2 + n^3 + n^4 $. [/et_pb_tab][et_pb_tab title="Hint 2" _builder_version="4.0" hover_enabled="0"]The idea is that we will try to bound the $ 1 + n + n^2 + n^3 + n^4 $ in between some squares and from that we will try to estimate the values of m in terms of n. Observe that $(2m)^2=4n^4+4n^3+4n^2+4n+4.$ Now, can you form squares from the right side? If not can you bound it by two squares? [/et_pb_tab][et_pb_tab title="Hint 3" _builder_version="4.0" hover_enabled="0"]First of all to form, you take the max terms $4n^4 = (2n)^2 $. So, that term must be included in the square. Also, try to find a, b, c such that $ (2n^2 + an + b)^2$ can be made greater or lesser the given expression. Observe that you will get the following. $(2n^2+n)^2<4n^4+4n^3+4n^2+4n+4<(2n^2+n+2)^2$ Now, guess that $(2n^2+n)^2<(2m)^2<(2n^2+n+2)^2$ So, what we get the relationship of m and n? [/et_pb_tab][et_pb_tab title="Hint 4" _builder_version="4.0" hover_enabled="0"]We get that $(2m) = (2n^2 + n + 1) $. Hence, $(2m)^2=(2n^2+n+1)^2 \Leftrightarrow 4n^4+4n^3+4n^2+4n+4=(2n^2+n+1)^2.$ Observe that, this results in a lot of cancellation of terms and we are left with: $n^2-2n-3=0.$ This gives the solution (m,n) = (11, 3) [/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, Korea Junior MO 2015, Problem 7

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" hover_enabled="0" box_shadow_style="preset2"]For a polynomial $f(x)$ with integer coefficients and degree no less than $1$ , prove that there are infinitely many primes $p$ which satisfies the following. There exists an integer $n$ such that $f(n) \not= 0$ and $|f(n)|$ is a multiple of $p$ . [/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="off" _builder_version="4.0" hover_enabled="0"]Korea Junior MO Problem 7 [/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="on"]Elementary Number Theory by David Burton [/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"]Well, remember the proof that the set of prime numbers is infinite? We started with the assumption that let there be a finite number of prime numbers and then reached a contradiction that there needs to be another extra prime number given that set. Hence, the set of prime numbers is infinite. This problem is also famously known as Schur's Theorem. Observe that the problem can be restated as every nonconstant polynomial p(x) with integer coefficients if S is the set of all nonzero values,

{\begin{Bmatrix}p(n)\neq 0:n\in {\mathbb {N}}\end{Bmatrix}}

then the set of primes that divide some member of S is infinite. Let us start by assuming that the set is indeed finite. Let $A$ this set of primes $p$ such that $\exists n$ such than $f(n)\ne 0$ and $p|f(n)$ . Let |A| be finite.
[/et_pb_tab][et_pb_tab title="Hint 2" _builder_version="4.0" hover_enabled="0"]If $f(0)=0$ the result is immediate since $p|f(p^n)$ $\forall p$ (just choose $n$ such that $f(p^n)\ne 0$ and so any prime $p\in A$ . Now let's take the case when f(0) is non-zero. Let's take $ f(x) = a_n.x^n + ... a_1.x + f(0)$. Now, $ f(c.f(0)) = a_n.{c.f(0)}^n + ... a_1.f(0) + f(0) = f(0).( a_n.c.{cf(0)}^{n-1} + ... + a_2.c^2.f(0) + a_1.c + 1 )$. Can you give some appropiate c to show that another prime must exist? [/et_pb_tab][et_pb_tab title="Hint 3" _builder_version="4.0" hover_enabled="0"]Take c = product of all the primes in A. Prove that it implies some other prime must exist which is not in A. [/et_pb_tab][et_pb_tab title="Hint 4" _builder_version="4.0" hover_enabled="0"]Now, $ f(c.f(0)) = a_n.{c.f(0)}^n + ... a_1.f(0) + f(0) = f(0).( a_n.c.{cf(0)}^{n-1} + ... + a_2.c^2.f(0) + a_1.c + 1 )$. Observe that if we take c as mentioned then, i.e. c = product of all the primes in A. Then all f(c.f(0)) must be coprime to all the primes in A. Therefore, it must have a prime factor other than those in A. Hence, a contradiction in the finiteness in A. 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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Inequality, Israel MO 2018, 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"]Determine the minimal and maximal values the expression $\frac{|a+b|+|b+c|+|c+a|}{|a|+|b|+|c|}$ can take, where $a,b,c$ are real numbers.[/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"]Israel MO 2018, Problem 3[/et_pb_accordion_item][et_pb_accordion_item title="Topic" _builder_version="4.0" open="off"]Algebra, Inequality[/et_pb_accordion_item][et_pb_accordion_item title="Difficulty Level" _builder_version="4.0" open="off"]6/10[/et_pb_accordion_item][et_pb_accordion_item title="Suggested Book" _builder_version="4.0" open="off"]Excursion in Mathematics by Bhaskarcharya 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"][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"]Given the expressions, what inequality comes to your mind first? The triangle inequality right? |x| + |y| $ \geq $ |x+y|. Can you use this inequality to get a maximum bound?[/et_pb_tab][et_pb_tab title="Hint 2" _builder_version="4.0"]Maximum Bound: Observe that the maximum bound is got by the triangle inequality as explained. $|a+b| \le |a|+|b|$
$|b+c| \le |b|+|c|$
$|a+c| \le |a|+|c|$ We get, $\frac{|a+b|+|b+c|+|c+a|}{|a|+|b|+|c|} \le\frac{|a|+|b|+|b|+|c|+|a|+|c|}{|a|+|b|+|c|}=2$ Never forget to mention the equality case: a = b = c is the equality case. [/et_pb_tab][et_pb_tab title="Hint 3" _builder_version="4.0"]What about the minimum inequality? The idea is that you can observe that keeping the denominator constant, can you reduce the numerator. Let's take the case of the |a| = |b| = |c| = 1. Now, the expression is maximized when a = b = c = 1 or -1. So, obviously one must be positive or two must be negative or vice-versa. In either case, we get $ \frac{2}{3} $. Okay, then maybe we need to deal with the signs and stuff to get a hold on the minimum. Let's fix the signs of a,b,c then, we can break the bonds of the modulus. Let's proceed to the next hint.

[/et_pb_tab][et_pb_tab title="Hint 4" _builder_version="4.0"]Let $a$ , $b$ , and $c$ be arbitrary real numbers, not all of them equal $0$ . By flipping signs, we can assume that at least two of $a$ , $b$ , and $c$ are non-negative. Actually, without loss of generality, we can assume that $a, b\geq 0$ . $ 3|a + b| + 3|b + c| + 3|c + a| \geq 3a + 3b + 3|b+c| + 3|a+c| \geq 2(a+b) + (a + |a + c|) + (b + |b + c|) $ $ = 2(|a| + |b|) + (|-a| + |a + c|) + (|-b| + |b + c|) \geq 2(|a| + |b|) + |c| + |c| = 2|a| + 2|b| + 2|c| $ We have proved that the minimum possible value of $\frac{|a+b|+|b+c|+|c+a|}{|a|+|b|+|c|}$ is $\frac{2}{3}$ . The minimum is $\frac{2}{3}$ , which is attained for $a = b = 1$ , $c = -1$ .

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Number Theory, Greece MO 2019, Problem 3

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[/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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Geometry, Israel 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|8px|20px||" box_shadow_style="preset2"]Six congruent isosceles triangles have been put together as described in the picture below. Prove that points M, F, C lie on one line.

[/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"]Israel MO 2019 Problem 3[/et_pb_accordion_item][et_pb_accordion_item title="Topic" _builder_version="4.0" open="off"]Geometry[/et_pb_accordion_item][et_pb_accordion_item title="Difficulty Level" _builder_version="4.0" open="off"]6/10[/et_pb_accordion_item][et_pb_accordion_item title="Suggested Book" _builder_version="4.0" open="off"]Challenges and Thrills of PreCollege 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"]You need to show that M, F, C lie on a straight line. Observe that it can be shown that they are collinear if we can show that $ \angle EFM = \angle CFD $.

We will now proceed towards proving in this direction.[/et_pb_tab][et_pb_tab title="Hint 2" _builder_version="4.0"]Let's investigate the triangle FDC. Observe that EF = AD and AD = AC. This results in the fact that FDC is isosceles and $ \angle FDC = \pi - \angle EDA $.

[/et_pb_tab][et_pb_tab title="Hint 3" _builder_version="4.0"]Now, we will try to infer something about triangle MEF. Observe that KM || HI as $ \angle MKJ = \angle KJH$. Hence KHIM must be a parallelogram. Hence, KH || MI. Also, $ \angle KHI =\angle HIG $. Hence, KH || EG. Hence, it implies from KH || MI and KH || EG, that M,E,I,G are collinear.

[/et_pb_tab][et_pb_tab title="Hint 4" _builder_version="4.0"]Now, this diagram ends it all. Observe that MI = KH. Also, EI = LJ. Hence, ME = KL = EP. Hence, MEF is isosceles. Also, $ \angle MEF = \pi - \angle GEF = \pi - \angle EDA = \angle FDC $. Hence, triangle MEF is similar to triangle FDC. This implies that $ \angle EFM = \angle DFC$.

QED

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