The discussion focuses on proving that the sequence n! + n^p - n + 2 contains at most finitely many squares for prime values of p = 3, 5 (mod 8). The solution approach involves using Fermat's Little Theorem, which leads to the simplification n! + n^p - n + 2 ≡ n! + 2 (mod p). It is established that for n ≥ p, n! ≡ 0 (mod p), resulting in the expression being equivalent to 2 (mod p). The key conclusion is that if 2 is not a quadratic residue modulo p, then the expression cannot yield a square for n ≥ p.
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Funky1981 said: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 said: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.