Number Theory - Italy MO 2019, Problem 2

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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 $n.$

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Start with hints

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[/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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Problem Solving Marathon Week1 Solution

Problem Solving Marathon Week1 Solution is the effortless attempt from Cheenta's existing student as well as from the end of mentor. Question, rules and hints are given here.

Level 0

Q.1 Which of the following is equal to $latex 1 + \frac{1}{1+\frac{1}{1+1}}$?

This solution is proposed by Swetaabh Mishra from Thousand Flowers. $latex 1 + \frac{1}{1+\frac{1}{1+1}} = 1 + \frac{1}{1+\frac{1}{2}}$
$latex =1 + \frac{1}{\frac{3}{2}}= 1+\frac{2}{3}=1\frac{2}{3} $

Answer of (Q.2) This solution is using hints.
If we pair up the elements of $latex X$ it will look like $latex (10,100)(12,98),(14,96),.....(54,56)$. Now sum of the each pair is $latex 110$. Number of pair $latex =\frac{Number of terms}{2} =\frac{46}{2}$. so $latex X$ will be equal to Number of pair $latex \times 110 =\frac{46}{2} \times 110=2530$ , similarly $latex Y$ will be $latex \frac{46}{2} \times 114=2622$.
So, $latex Y-X$ will be $latex 92$.

Level 1

Q.1 Find all positive integers $latex n$ such that $latex n^2+1$ is divisible by $latex n+1$.

Since, $latex n^2+1$ can be written as $latex n(n+1)-(n-1)$
We can say that if $latex n+1|n^2+1$ then $latex n+1|n-1$. At a glance, it's look like impossible to get any positive integer. If $latex n-1=0$ then it is possible. So there is only one such positive integer $latex n=1$.

Also Visit: Cheenta Olympiad Program

Q.2 Two geometric sequences $latex a_1, a_2, a_3, \ldots$ and $latex b_1, b_2, b_3, \ldots$ have the same common ratio, with $latex a_1 = 27$, $latex b_1=99$, and $latex a_{15}=b_{11}$. Find $latex a_9$.
Example of Geometric Sequence $latex 2,4,8,16$, here common ratio is $latex 2$.

This solution is proposed by Saikrish Kailash from Thousand Flowers.
Two geometric sequence have same common ratio, let it is $latex r$.
Now $latex a_9=27\times r^{(9-1)}=27r^8$. From the given condition $latex a_{15}=b_{11} $. Which is equivalent to $latex 27r^{14}=99r^{10} \Rightarrow r^4=\frac{11}{3}$. Now put the value of $latex r^4$ in $latex a_9=27r^8=27(r^4)^2=363$.

Level 2

Q.1 Let $latex m$, $latex n$, $latex p$ be real numbers such that $latex m^2 + n^2 + p^2 - 2mnp = 1$ . Prove that $latex (1+m)(1+n)(1+p) \leq 4 + 4mnp$.

This solution is proposed by Sampreety Pillai from Early Bird Math Olypmiad Group.
We konw $latex (m-n)^2 \geq 0$ which imply $latex m^2+n^2 \geq 2mn$. similarly $latex n^2+p^2 \geq 2np$, $latex p^2+m^2 \geq 2mp$. Also $latex m^2+1 \geq 2m$, $latex n^2+1 \geq 2n$ and $latex p^2+1 \geq 2p$. Adding these six inequalities, we get $latex 3(m^2+n^2+p^2+1) \geq 2(mn+mp+np+m+n+p)$. by the hypothesis $latex m^2+n^2+p^2=2mnp+1$.
So $latex 3(mnp+1) \geq 2(mn+mp+np+m+n+p)$. Add $latex mnp+1$ both side of last inequality.
Also you can do another type of proof using Cauchy-Schwarz inequality. For that click here.