Prove that $11^{2n}+5^{2n+1}-6$ is divisible with $24$ for $n∈ℤ^+$

Question

I've been trying to solve it by using modulo; $11^{2n}+5^{2n+1}-6≡ (11^2 mod24)^n + 5*(5^2mod24)^n-6 = 1^n + (5*1^n)-6 = 0$

Is this the right way to tackle the problem? I am not certain if I am placing the "$mod$" marks at the right places.

You can use \pmod or \bmod or just \mod: $2\pmod{3}, 2\bmod{3}, 2\mod 3$. — Daniel W. Farlow, May 10 '16 at 17:26
$5^{2n+1}\equiv5\ (mod\ 24)$ and $11^{2n}\equiv1\ (mod\ 24)$. $5+1-6=0$ — Mastrem, May 10 '16 at 17:27

score 1 · Accepted Answer · answered May 10 '16 at 17:58

1

$$11^{2n}+5^{2n+1}-6=(5\cdot24+1)^n+5(24+1)^n-6\equiv 1+5-6\equiv0\pmod{24}$$

answered May 10 '16 at 17:58

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