Can't find n (algebraic manipulation)

I‘m trying to understand a proof on my textbook that says $$sqrt {n − 1} +sqrt {n + 1}$$ is irrational, where $$n in mathbb{N}$$. However I can’t seem to show that
$$n^2 – 1 = frac{(frac{p^2}{q^2} – 2n)^2}{4} to n = frac{(p^4+4q^4)}{4p^2q^2}$$,
where $$p,q in mathbb{Z}$$，and the gcd of $$p,q$$ is 1. I feel really stupid, is there something I don’t see?

The furthest I got is
$$n = frac{sqrt{2p^4+5p^4-2p^2q^2}-(p^2-q^2)}{2q^2}$$

Mathematics Asked by Heng Wei on December 30, 2020

$$n^2 - 1 = frac{left(dfrac{p^2}{q^2} - 2nright)^2}{4}= frac{dfrac{p^4}{q^4} + 4n^2-4ndfrac{p^2}{q^2} }{4}=dfrac {p^4}{4q^4}+n^2-ndfrac{p^2}{q^2}.$$

Subtract $$n^2$$ from both sides to get $$-1=dfrac {p^4}{4q^4} -ndfrac{p^2}{q^2} ,$$

which is $$ndfrac{p^2}{q^2}=1+dfrac {p^4}{4q^4}.$$

Can you take it from here (solve for $$n$$)?

Correct answer by J. W. Tanner on December 30, 2020

If $$n^2 - 1 = frac{(frac{p^2}{q^2} - 2n)^2}{4}$$ then $$n^2 = frac{(frac{p^2}{q^2} - 2n)^2+4q^2}{4}$$ then $$n^2 = frac{(p^2 - 2q^2n)^2+4q^2}{4q^2}$$ then $$4q^2n^2 = (p^2 - 2q^2n)^2+4q^2$$ then $$4q^2n^2 = 4q^4n^2-4p^2q^2n+p^4+4q^2$$ so $$0=-4p^2q^2n+p^4+4q^2$$ and finally $$n=frac{p^4+4q^4}{4p^2q^2}$$

Answered by Measure me on December 30, 2020

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