You cannot use any theorems involving modular arithmetic.

1.) Prove the theorem that "Let \(a,b \in \mathbb{Z}\) such that \(a \neq 0\) or \(b \neq 0\), we get that \(\exists x,y \in \mathbb{Z}, (a,b) = ax+by\)"

Note: \((a,b)\) is the greatest common divisor.

2.) Let \(p\) be a prime number. Prove that \(\exists k \in \mathbb{Z}, (p-1)! + 1 = p^{k}\) if and only if \(p = 2,3,5\).

3.) Prove that if \(p\) and \(8p^{2}+1\) are prime numbers, then \(8p^{2}+2p+1\) is also prime number.

4.) Prove that there exists infinitely many positive integer \(n\) such that \(10^{n}+3\) is composite.

5.) Find all such primes \(p\) and positive numbers \(n\) such that \(n^{p} + 3^{p}\) are perfect squares.

This is the part of Thailand 1st round math POSN problems.

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## Comments

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TopNewestFor 3, 8p^2 +1 is a multiple of 3 if p isn't equal to3 . so, p = 3 and hence forth, it becomes trivial

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1.) is just the Euclidean algorithm, but I can't just freaking prove it nuuuu.

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For 4, if n is of form 6k + 4 , then 10^n + 3 is a multiple of 7, hence composite

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\(Problem \enspace 3:\)

If \(p=2, \enspace 8p^2+1=8(2)^2+1=33 ;\)Not prime

If \(p=3, \enspace 8p^2+1=8(3)^2+1=73 \enspace\) and \(\enspace 8p^2+2p+1=8(3)^2+2(3)+1=79 ;\) both are primes.

Other prime numbers can be written as \(6k\pm1\).

Then, \(8p^2+1 = 8(6k\pm1)^2+1=8(36k^2\pm12k+1)+1 = 3(96k^2\pm32k+3)\) ; which can't be a prime.

So only for \(p=3\), \(8p^2+1\) and \(8p^2+2p+1\) both become prime.

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