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.

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## Comments

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TopNewest1) 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

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

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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)

Adding (1) & (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.

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

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