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Let $p=4k+1,$$p$ is a prime numbe,and $ k\in\mathbb{Z}$. Prove the existence of positive integer $a_1,a_2,\ldots,a_k$ and $b_1,b_2,\ldots,b_k$. $$ p=\frac{(a^{2}_{1}+1)(a^{2}_{2}+1)\cdots(a^{2}_{k}+1)}{(b_{1}^{2}+1)(b_{2}^{2}+1)\cdots(b_{k}^{2}+1)}. $$ My friend trying to prove the following for a long time,and I cann't solve it,I hope you can help me. I would be extremely happy if somebody could help me with this!

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Let $n \in {\mathbb N}$, and $n > 3$. Prove the non-existence of positive integer $x, y, z$. $$x^n + y^n = z^n$$ My friend trying to prove the following for a long time,and I cann't solve it,I hope you can help me. I would be extremely happy if somebody could help me with this! It is definitely not the question for stackexchange. –  penartur Apr 24 '12 at 5:06
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Where did the problem come from? Is there any reason to believe it holds? Are there any restrictions on $a_i$ and $b_i$? Context? –  Arturo Magidin Apr 24 '12 at 5:07
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@penartur Why is this question definitely not a question for stackexchange? Where else would you go? –  draks ... Apr 24 '12 at 6:27
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@penartur (1) This site is for all math questions, simple or difficult. (2) Even if difficult questions were disallowed, generally there is no way to know in advance how difficult a solution will be. Teachers sometimes pose difficult (even unsolved) problems as exercises without any warning to students. Occasionally they are solved. Incorrect preconceptions about the difficulty of problems often times prevent one from discovering simple, beautiful solutions. –  Bill Dubuque Apr 24 '12 at 14:23
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@penartur:Thank you for your concern, I am a Chinese student, my English is terrible.I think I will try to overcome language difficulties.It will be a pleasant thing ,exchange Mathematics with you.thank you! –  Daoyi Peng Apr 25 '12 at 10:02

1 Answer 1

Maybe something like this works. There's an integer $a_1$ such that $a_1^2+1=pm_1$ with $m_1\lt p$. There's an integer $b_1$ such that $b_1^2+1=m_1n_2$ with $n_2\lt m_1$. There's an integer $a_2$ with $a_2^2+1=n_2m_2$ with $m_2\lt n_2$. Etc. The sequence $p\gt m_1\gt n_2\gt m_2\dots$ must terminate after at most $k$ steps (since each $m_i$ and $n_i$ is 1 mod 4).

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