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Are there any such $n,q\in\mathbb{Q}\setminus\{0,1\}$ so that $n^2+q^2=1$?
How could one prove their (in)existence?

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2 Answers

up vote 12 down vote accepted

Consider the Pythagorean triplets $(a,b,c)$ and set $n = \pm \dfrac{a}c$ and $q = \pm \dfrac{b}c$. Hence, there are infinitely many solutions to your problem.

All the Pythagorean triplets are generated uniquely by the following formula: $$a = k(m^2-n^2), b = k(2mn) \text{ and } c=k(m^2+n^2)$$ where $m,n,k \in \mathbb{Z}^+$, such that $m > n$ are of opposite parity and are also relatively prime.


To show that these are all the solutions, let $n = \dfrac{n_1}{n_2}$ and $p = \dfrac{p_1}{p_2}$, where $(n_1,n_2) = 1 = (p_1,p_2)$. We then need $$\left(\dfrac{n_1}{n_2} \right)^2 + \left(\dfrac{p_1}{p_2} \right)^2 = 1$$ which on rearranging gives us $$(n_1p_2)^2 + (n_2 p_1)^2 = (n_2 p_2)^2$$ This gives us $p_2^2 \vert (n_2 p_1)^2$ and since $(p_1,p_2) = 1$, we get that $p_2^2 \vert n_2^2$. Arguing the other way around, we get that $n_2^2 \vert p_2^2$ and hence, we have $n_2^2 = p_2^2$. We can always push the signs to the numerator in a fraction and hence we can conclude that $n_2 = p_2$ and say both are equal to $q \in \mathbb{Z}^+$.

Hence, all rational solutions we are after are of the form $\dfrac{n_1}{q}$ and $\dfrac{p_1}q$, where $(n_1,q) = 1 = (p_1,q)$ and hence this is equivalent to finding relatively prime Pythagorean triplets, which are given by $$a = (m^2-n^2), b = (2mn) \text{ and } c=(m^2+n^2)$$ where $m,n \in \mathbb{Z}^+$, such that $m > n$ are of opposite parity and are also relatively prime.

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It may be considered worthwhile to add a proof that these are all solutions. –  Lord_Farin Apr 9 '13 at 22:10
@Lord_Farin Added. –  user17762 Apr 9 '13 at 22:57
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If $t\in\mathbb R$, then $$(t^2-1)^2+4t^2=(t^2+1)^2,$$ therefore we can let $$\tag1n=\frac{t^2-1}{t^2+1}, q=\frac{2t}{t^2+1}.$$ (Rational parametrization of the circle). Of course, the easiest way to guarantee $n,q\in\mathbb Q$ is to start with $t\in\mathbb Q$. On the other hand, if $n,q$ are rational and $q\ne 0$, then $$ \frac 1q-\frac nq=\frac{t^2+1}{2t}-\frac{t^2-1}{2t}=\frac1t$$ is also rational, i.e. $(1)$ with $t\in\mathbb Q$ are in fact all solutions (and we avoid $t=0$, $t=1$ and the limit cases $t=\pm\infty$ as these would lead to $n,q\in\{0,1,-1\}$).

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