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 (54a−1,52b−1) using Euclidean algorithm.
2.) (same as last year) Prove that if p and 8p2+1 are prime numbers, then 8p2+2p+1 is also prime number.
3.) Find all positive integers n such that
12557+22557+⋯+n2557+(n+1)2557
is a composite number.
4.) Let n∈N. Prove that 22n+1+22n+1 has at least n+1 distinct prime factors.
5.) Let a,b,c∈N such that (a,b,c)=1 and a1+b1=c1. Prove that a+b is a perfect square.
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Comments
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
Q.5 Is from RMO-1992(Q.2)
1)choosing smallest no. of b ( b=1) ...... 52b-1 = 52-1 = 24
similarly choosing smallest no. of a( a=2) ..... 54a-1 = 58- 1 (divides 24 as follows:)
52≡1(mod24)
=> 58≡1(mod24)
So gcd = 24
3) when no. divided by 2 leaves remainder 0 . no. is composite.
1≡1(mod2)
11557≡1(mod2)...... (1)
3≡−1(mod2)
31557≡−1(mod2).......(2)
Adding (1) & (2) ....
11557+31557≡0(mod2)
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.