Number Theory, France IMO TST 2012, Problem 3

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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"]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_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"]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, 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 - Germany MO 2019, Problem 4

Understand the problem

Show that for each non-negative integer $n$ there are unique non-negative integers $x$ and $y$ such that we have
$n=\frac{(x+y)^2+3x+y}{2}.$

[/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"]Germany MO, 2019, Problem 4 [/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"]4/10 [/et_pb_accordion_item][et_pb_accordion_item title="Suggested Book" _builder_version="4.0" hover_enabled="0" open="on"]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"]If you observe the expression $ \frac{(x+y)^2 + 3x + y}{2} $. It has two free variables. Now, if let's play with the expression to make it a little more symmetric and easy to work with and make some meaning out of it. Try to remove the asymmetry of x and write it in a much more symmetric form in x and y.

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

Observe that $ \frac{(x+y)^2 + 3x + y}{2} = \frac{(x+y)^2 + (x+y)}{2} + x$ is the expression what we get after bringing in the symmetry. Now, factorize it and see what we are looking for is $ n = \frac{(x+y)(x+y+1)}{2} + x$. Can you guess anything about the expression $ \frac{(x+y)(x+y+1)}{2} $.

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

$ \frac{(k)(k+1)}{2} $ is the sum of the first k natural numbers. So, now the idea is that somehow you are taking the first k natural numbers and adding another number x to it to make any number. Can you get the final logic? [/et_pb_tab][et_pb_tab title="Hint 4" _builder_version="4.0" hover_enabled="0"]Now, observe that sum of the first k numbers is increasing with k. Now, take any number say 17. Now observe that 17 lies between 15 and 21. This means that 17 = 15 + 2 = ( 1 +2 + 3 +4 + 5) + 2. So, this is the idea that k = x+y is the largest number such that n is greater than or equal to the ( 1 + 2 + ... + k ) and observe that we may need something like 2 in case of n = 17. Call that x and obviously $ k \geq x $. Hence, define y = k - x. So, for n = 33 = ( 1 + 2 + ... + 7 ) + 5. Hence k = x + y = 7, x = 5, y = 2. This is the whole idea. 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 - Italy MO 2019, Problem 2

Understand the problem

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Let $p,q$ be prime numbers $.$ Prove that if $p+q^2$ is a perfect square $,$ then $p^2+q^n$ is not a perfect square for any positive integer $n.$

[/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" hover_enabled="0"][et_pb_accordion_item title="Source of the problem" open="on" _builder_version="4.0" hover_enabled="0"]Italy MO 2019, Problem 2 [/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"]5/10 [/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 first write the condition and use the idea that p and q are prime. Let $a \in \mathbb{N}$ such that $p + q^2 = a^2$ => $p = a^2 - q^2 = (a - q)(a + q)$ . As $p$ is a prime and $a + q > a - q$ , then $a - q = 1$ . We must have $p = 2q + 1$ . Now, try to use this condition to rewrite the expression we are interested in. [/et_pb_tab][et_pb_tab title="Hint 2" _builder_version="4.0" hover_enabled="0"]p = 2q+1. Suppose that there exists $b \in \mathbb{N}$ such that for some positive integer $n$ , then
$p^2 + q^n = b^2$ => $q^n = b^2 - p^2$ => $q^n = (b - p)(b + p)$ . It implies that both (b-p) and (b+p) must be powers of q. Hence gcd(b-p, b+p) = (b-p). b-p | b+p => b-p | 2p = 4q + 2. As, b-p is a power of q, it implies that b-p = 1 and q can be a prime or q = 2 and b-p = 2 as gcd (b-p, 4q+2) = 1 if q is an odd prime or 2 if q = 2. Thus, we get that ( b-p = 1 ) or ( b-p =2, q = 2 ) and p = 2q+1. [/et_pb_tab][et_pb_tab title="Hint 3" _builder_version="4.0" hover_enabled="0"]Case 1 ( b-p = 1 ): b -p = 1, (b - p) = 1, (b+p) = $q^n = 1 + 2p = 3 + 4q$. Now, you see that LHS will grow much more fast as q increases than the RHS. Mathematically, $ q^n \geq q^2 \geq 4q+3 $ as $ q \geq 5, n \geq 2 $. So, the only possibility for q is q = 2, 3, 5. But it has no solution as you can see. [/et_pb_tab][et_pb_tab title="Hint 4" _builder_version="4.0" hover_enabled="0"]Case 2 ( q = 2 ): It implies p = 2q+1 = 5. $ 2^n = (b-5)(b+5) $. It implies that the both b-5 and b+5 must be powers of 2. Let $ b-5 = 2^a \leq b+5 = 2^b \rightarrow 2^b - 2^a = 10 \rightarrow a = 1 and 2^(b-1) - 1 = 5 $. This has no solution.

QED.

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Extremal Principle : I.S.I Entrance 2013 problem 4

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Understand the problem

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exactly once and each game had a winner. After all the games, each player
listed the names of all the players she defeated as well as the names of all the
players defeated by the players defeated by her. For instance, if A defeats B
and B defeats C, then in the list of A both B and C are included. Prove that
at least one player listed the names of all other players.

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I.S.I. (Indian Statistical Institute) B.Stat/B.Math Entrance Examination 2013. Subjective Problem no. 4.

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7.5 out of 10

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Problem Solving Strategies by Engel

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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="3.22.4"]

Do you know what is Well-ordering principle ? it says that Every nonempty set A of nonnegative integers has a minimal element and a maximal element , which need not to be unique . now some time this simple property help us to get very nice solution to a problem , can you some how apply this property .

[/et_pb_tab][et_pb_tab title="Hint 2" _builder_version="3.22.4"]

Use the method of contradiction , first of all assume that there is no player which have the given property . Now try to use the property of hint 1 .

[/et_pb_tab][et_pb_tab title="Hint 3" _builder_version="3.22.4"]

If there is no such list , A's list has the maximum no. of players Now , if A does not have the certain property then there exist another another player B , who has won against A . Now B's list contain the name of A [ by the 1st condition ] and all the names of the players defeated by A [ by the 2nd condition]

[/et_pb_tab][et_pb_tab title="Hint 4" _builder_version="3.22.4"]

Now , can you find out some contradiction , yes exactly ..... B's list contain more number of element than A So, A's list must have the certain property .

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Connected Program at Cheenta

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Indian Statistical Institute and Chennai Mathematical Institute offer challenging bachelor’s program for gifted students. These courses are B.Stat and B.Math program in I.S.I., B.Sc. Math in C.M.I.

The entrances to these programs are far more challenging than usual engineering entrances. Cheenta offers an intense, problem-driven program for these two entrances.

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