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What I have done so far:

  • $n$ must be even
  • $(n-1)(n+1)|2^{n!}-1$
  • because $n-1$ and $n+1$ are odd, they are relatively prime
  • so both $n-1$ and $n+1$ should divide $2^{n!}-1$

Then I get lost in how I should handle it. Could anyone help me with that?

J.G.
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WANYUI
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  • A better way to put it is that $n^2-1|2^{n!}-1$ iff $n\pm1$ both divide $2^{n!}-1$. – J.G. Nov 18 '20 at 15:56
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    See https://math.stackexchange.com/questions/2126531/prove-that-for-any-even-positive-integer-n-n2-1-mid-2n-1, or https://math.stackexchange.com/questions/1019141/prove-that-big-n2-1-big-mid-big-2n-1-big. It seems that this is true for all even $n$. – player3236 Nov 18 '20 at 16:00
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    Thank you guys for your help. The solution seems clear;) – WANYUI Nov 18 '20 at 17:26

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