Question

prove that n^7 plus 7 is not a perfect square number

Answer 1

The first thing I thought of was a proof by contradiction but it was a bit too much. This proof is a proof by induction and it uses calculus principles:

Base case: n = 1. For n = 1, the expression evaluates to 8 which is not a perfect prime.
Inductive step: Assume that for n = k, that \[k ^{7} + 7\]is not prime.
Substituting k + 1 into k, we get:
\[(k+1)^{7} + 7 = k ^{7} + 8 + 7k(k+1)(k ^{2}+k + 1)^{2}\]
This expression is a perfect square if and only if it equals
\[b ^{2}\]
for some natural number b.
However, it is evident that these two curves will NEVER intersect because the derivative of the first is always greater than the second. Therefore by induction,
\[n ^{7} + 7\]
is never a perfect square.

Answer 2

If you need the proof that the curves never intersect via the derivative, I'll gladly give you that part, too.

Answer 3

Sorry, I reread that proof. It is perfect SQUARES not perfect PRIME.