Proving Finite Squares in n!+n^p-n+2 for p=3,5 (mod8)

[Funky1981]

Homework Statement

How to prove the following:

Let p be a prime p=3,5 (mod8). Show that the
sequence n!+n^p-n+2 contains at most finitely many squares.

Should I build a contardiction or prove it directly? I really need some help


2. The attempt at a solution
Use Fermats Little we have n^p-n=0 (mod p) then n!+n^p-n+2=n!+2(mod p)
how should i keep going??

 

[Dick]

Homework Statement

How to prove the following:

Let p be a prime p=3,5 (mod8). Show that the
sequence n!+n^p-n+2 contains at most finitely many squares.

Should I build a contardiction or prove it directly? I really need some help 2. The attempt at a solution
Use Fermats Little we have n^p-n=0 (mod p) then n!+n^p-n+2=n!+2(mod p)
how should i keep going??

To keep going you should notice that if n>=p then n!=0(mod p) so n!+n^p-n+2=2(mod p). Now if you can show 2 is NOT a quadratic residue of p, then for n>=p, that expression cannot be a square.

 

[Funky1981]

To keep going you should notice that if n>=p then n!=0(mod p) so n!+n^p-n+2=2(mod p). Now if you can show 2 is NOT a quadratic residue of p, then for n>=p, that expression cannot be a square.

Thank u so much!