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I've tried using Pythagoras theorem but it always results in a silly answer like $n=n^2$ or something.

I'm nearly 100% sure this is done with Pythagoras but I'm not sure which way to do it

iadvd
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Tommy101
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  • $(n^2-1)^2=n^4-2n^2+1$ and $(n^2+1)^2=n^4+2n^2+1$ and $(2n)^2=4n^2$ – robjohn Jun 10 '15 at 07:54
  • Sounds like you make a mistake in your algebra. You should have ended with something like $n^2 = n^2$. If you show your work, you will probably find your error yourself. – marty cohen Jul 04 '15 at 17:06

2 Answers2

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Hint: $$ 4n^2 = (n^4 + 2n^2 +1) - (n^4 - 2n^2 + 1)$$

corindo
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This is a particular case of Pythagorean triples. All such triples $(a,b,c)$ with $a^2+b^2=c^2$ can be generated by two integers $m,n$ as: $$ a=n^2-m^2,\qquad b=2mn,\qquad c=m^2+n^2 $$ (the identy $a^2+b^2=c^2$ is easely verified). In your case we have $m=1$.

Emilio Novati
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