Number Theory (1st math Thailand POSN 2014)

You cannot use any theorems involving modular arithmetic.

1.) Prove the theorem that "Let a,bZa,b \in \mathbb{Z} such that a0a \neq 0 or b0b \neq 0, we get that x,yZ,(a,b)=ax+by\exists x,y \in \mathbb{Z}, (a,b) = ax+by"

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

2.) Let pp be a prime number. Prove that kZ,(p1)!+1=pk\exists k \in \mathbb{Z}, (p-1)! + 1 = p^{k} if and only if p=2,3,5p = 2,3,5.

3.) Prove that if pp and 8p2+18p^{2}+1 are prime numbers, then 8p2+2p+18p^{2}+2p+1 is also prime number.

4.) Prove that there exists infinitely many positive integer nn such that 10n+310^{n}+3 is composite.

5.) Find all such primes pp and positive numbers nn such that np+3pn^{p} + 3^{p} are perfect squares.

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

Note by Samuraiwarm Tsunayoshi
5 years, 1 month ago

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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

A Former Brilliant Member - 5 years, 1 month ago

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

Samuraiwarm Tsunayoshi - 5 years, 1 month 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

A Former Brilliant Member - 5 years, 1 month ago

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Problem3:Problem \enspace 3:

If p=2,8p2+1=8(2)2+1=33;p=2, \enspace 8p^2+1=8(2)^2+1=33 ;Not prime

If p=3,8p2+1=8(3)2+1=73p=3, \enspace 8p^2+1=8(3)^2+1=73 \enspace and 8p2+2p+1=8(3)2+2(3)+1=79;\enspace 8p^2+2p+1=8(3)^2+2(3)+1=79 ; both are primes.

Other prime numbers can be written as 6k±16k\pm1.

Then, 8p2+1=8(6k±1)2+1=8(36k2±12k+1)+1=3(96k2±32k+3)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=3p=3, 8p2+18p^2+1 and 8p2+2p+18p^2+2p+1 both become prime.

Fahim Shahriar Shakkhor - 5 years, 1 month ago

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