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Prove that $5^{125}-1$/ ($5^{25} - 1$) is composite

I have written $5^{125}-1$ as $(5^{25}-1)(5^{100}+5^{75}+5^{50}+5^{25}+1)$ but what should I do after this? Sorry about earlier mistake in question ,

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    $\begingroup$ Why do you need to factorize any further? You have found a number $k$ such that $(5^{25}-1)\cdot k = 5^{125}-1$ right? $\endgroup$ – Prahlad Vaidyanathan Dec 9 '13 at 7:22
  • $\begingroup$ Just to clarify, $5^{100}+5^{75}+5^{50}+5^{25}+1$ is an integer! $\endgroup$ – Laars Helenius Dec 9 '13 at 7:24
  • $\begingroup$ Well $5^{100}+5^{75}+5^{50}+5^{25}+1\equiv 1\pmod5$ and $5k+1$ is even if $k$ is odd. But $5^{99}+5^{74}+5^{49}+5^{24}$ is even! $\endgroup$ – Laars Helenius Dec 9 '13 at 7:34
  • $\begingroup$ @user112790. $5^{99}+5^{74}+5^{49}+5^{24}$ is even. $\endgroup$ – hrkrshnn Dec 9 '13 at 7:36
  • $\begingroup$ As soon as I posted it, I realized my error. All fixed now $\endgroup$ – Laars Helenius Dec 9 '13 at 7:44
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Let $x = 5^{25}$.

$\begin{align} 5^{125}-1 &= x^5-1\\ &=(x^4 +x^3 +x^2 + x + 1)(x-1) \\ &= (x^4 + 9x^2 + 1 + 6x^3 + 6x + 2x^2 - 5x^3 - 10x^2 - 5x)(x - 1)\\ &= ((x^2 + 3x + 1)^2 - 5x(x + 1)^2)(x - 1) \end{align}$

Put $x = 5^{25}$, just in the expression $5x$, you will get

$5^{125}-1=((x^2 + 3x + 1)^2 - (5^{13}(x + 1))^2)(x - 1)$

it is now of the form $(a^2-b^2)(x-1)$ where $a,b$ are integers and hence $(x^5-1)/(x-1)$ is a composite number.

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  • $\begingroup$ Note that you still need to check that the two factors of $a^2-b^2$ are not trivial. $\endgroup$ – Phira Dec 27 '13 at 0:20

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