# How to prove that?

Proof that for any prime number $q>3$? $$(qn-1)^q(2qn+1) -(qn+1)^q +2^q$$ gives remainder $0$ when divided by $q^3$ for some $n\in N$,$n\ge 3$

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Your problem is that you had us going by saying for all $n\in\mathbb{N}$ instead of for some $n$. Your bigger problem is that you didn't provide your background thoughts and math, thereby wasting others' time and energy. Be more transparent and you won't get downvoted like this next time. –  anon Nov 6 '11 at 22:27

The edited form of the question is still incorrect: if $q=5$, $n=2$, then $$(qn-1)^q(2qn+1) -(qn+1)^q+2^q=(9)^5(21)-(11)^5+2^5=1079010\equiv 10\not\equiv0\bmod 125.$$
The edited form of the question is still incorrect: if $q=5$, $n=4$, then $$(qn-1)^q(2qn+1) -(qn+1)^q+2^q=(19)^5(41)-(21)^5+2^5=97435990\equiv 115\not\equiv0\bmod 125.$$
$((5*3-1)^5(2*5*3+1)-(5*3+1)^5 +2^5)/2^5 = 488250\equiv 0 mod 125$ –  Chun-Yue Nov 6 '11 at 21:38