# Number Theory (Thailand Math POSN 1st elimination round 2014)

Write a full solution.

1.) Let $a$ be an even number and $b$ an odd number such that $(a,b) = 1$. Find the value of $\left(5^{4a}-1,5^{2b}-1\right)$ using Euclidean algorithm.

2.) (same as last year) Prove that if $p$ and $8p^{2}+1$ are prime numbers, then $8p^{2}+2p+1$ is also prime number.

3.) Find all positive integers $n$ such that

$1^{2557} + 2^{2557} + \dots + n^{2557} + (n+1)^{2557}$

is a composite number.

4.) Let $n \in \mathbb{N}$. Prove that $2^{2^{n+1}} + 2^{2^{n}} + 1$ has at least $n+1$ distinct prime factors.

5.) Let $a,b,c \in \mathbb{N}$ such that $(a,b,c) = 1$ and $\displaystyle \frac{1}{a}+\frac{1}{b} = \frac{1}{c}$. Prove that $a+b$ is a perfect square.

Check out all my notes and stuffs for more problems!

Thailand Math POSN 2013

Thailand Math POSN 2014

Note by Samuraiwarm Tsunayoshi
6 years, 6 months ago

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1) is obviously 24.2) obviously p=3, which gives a satisfying result(79 is prime).3) is always composite (check divisibility by n or n+1) 5) use SFFT to get (a-c)(b-c)=c^2.But since they are coprime, we get the only sol c^2+c+1,c+1, whose sum is (c+1)^2

- 6 years, 6 months ago

Q.5 Is from RMO-1992(Q.2)

- 6 years, 6 months ago

1)choosing smallest no. of b ( b=1) ...... $5^{2b}$-1 = $5^{2}$-1 = 24

similarly choosing smallest no. of a( a=2) ..... $5^{4a}$-1 = $5^{8}$- 1 (divides 24 as follows:)

$5^{2} \equiv 1 \pmod{24}$

=> $5^{8} \equiv 1 \pmod{24}$

So gcd = 24

3) when no. divided by 2 leaves remainder 0 . no. is composite.

$1 \equiv 1 \pmod{2}$

$1^{1557} \equiv 1 \pmod{2}$...... (1)

$3 \equiv -1 \pmod{2}$

$3^{1557} \equiv -1 \pmod{2}$.......(2)

$1^{1557}+ 3^{1557} \equiv 0 \pmod{2}$

all even to some power will be composite .

So , n= 3,7, 11 ..... should be the nos.

- 6 years, 5 months ago

Someone told me that no.3) every positive integers make the sum composite, but I don't know how to prove that.

- 6 years, 5 months ago