practice problems pending

Q1. If x^n + y^n is a prime number for some x,y > 1 then n is necessarily:
Options: Power of 2, Multiple of 4, Multiple of 3, Odd

S1.
Answer: Power of 2.
Proof by contradiction:

Assume that n has an odd factor k > 1.
n = m.k
x^n + y^n = (x^m)^k + (y^m)^k
A^k + B^k = (A+B)(A^k-1 -A^k-2.B ....+ B^k-1) for odd 'k'.
Here A = x^m and B = y^m
x,y > 1 m>=1 => A+B >= 4
And the other factor is also more than 1 since A^k + B^k > A + B since k > 1.
So A^k + B^k is not a prime.
So n can't have any odd factor.
So n has to be power of 2.
H.P.

For e.g.
n = 2
x = 2, y = 3
4 + 9 = 13


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