Can't find n (algebraic manipulation)

Question

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
$$n = frac{sqrt{2p^4+5p^4-2p^2q^2}-(p^2-q^2)}{2q^2}$$

J. W. Tanner · Accepted Answer

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

Measure me · Answer

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
$$n=frac{p^4+4q^4}{4p^2q^2}$$

Can't find n (algebraic manipulation)

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