# Number Theory (1st math Thailand POSN 2014)

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. Note by Samuraiwarm Tsunayoshi
6 years, 8 months ago

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## Comments

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

- 6 years, 8 months ago

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

- 6 years, 8 months ago

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

- 6 years, 8 months ago

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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.

- 6 years, 8 months ago

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