×

# 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
2 years ago

Sort by:

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 · 2 years ago

$$Problem \enspace 3:$$

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. · 2 years ago

For 4, if n is of form 6k + 4 , then 10^n + 3 is a multiple of 7, hence composite · 2 years ago