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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

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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 $$(qn-1)^q(2qn+1) -(qn+1)^q+2^q=(19)^5(41)-(21)^5+2^5=97435990\equiv 115\not\equiv0\bmod 125.$$

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Yes, sorry for that. I had problems formatting the message and deleted part of it. –  Chun-Yue Nov 6 '11 at 21:16
@Chun-Yue: The new version of the question is still incorrect. You should use Wolfram Alpha to test these statements for yourself, instead of asking a different version here every time. It's not appropriate to keep changing the question the way you are. –  Zev Chonoles Nov 6 '11 at 21:26
$((5*3-1)^5(2*5*3+1)-(5*3+1)^5 +2^5)/2^5 = 488250\equiv 0 mod 125$ –  Chun-Yue Nov 6 '11 at 21:38
No, it is your job to ask a question that does not have a trivial counterexample. –  Zev Chonoles Nov 6 '11 at 21:46
I am not sure how you came to believe those statements were equivalent. However, that statement does seem to be true, and I don't see immediately why. You should ask a new question with that statement, and be sure to explain your thoughts about the question, and where it is coming from - is it homework? your own conjecture? –  Zev Chonoles Nov 6 '11 at 22:11

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