# How to prove that? [closed]

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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## closed as off-topic by Jonas Meyer, Ittay Weiss, Grigory M, Najib Idrissi, dustinJan 18 at 18:36

This question appears to be off-topic. The users who voted to close gave this specific reason:

• "This question is missing context or other details: Please improve the question by providing additional context, which ideally includes your thoughts on the problem and any attempts you have made to solve it. This information helps others identify where you have difficulties and helps them write answers appropriate to your experience level." – Jonas Meyer, Ittay Weiss, Grigory M, Najib Idrissi, dustin
If this question can be reworded to fit the rules in the help center, please edit the question.

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