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Can you give two numbers $x,y\in\mathbb{Q}$ such that ${ x }^{ 4 }+{ y }^{ 2 }=1$?

I don't know if exists or not. I derived this equation questioning that if $\sin { \alpha } ={ x }^{ 2 }$ for $x\in \mathbb{Q}$ then for which $\alpha$, $\cos{ \alpha }=y$ and $y\in \mathbb{Q}$?

Edit: $x,y\neq0$

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Consider $\dfrac{p}{q}=x$ and $y=\dfrac{m}{n}$, such that $\gcd (p,q)=(m,n)=1$ and $n \neq 1$, $q \neq 1$

$(\dfrac{p}{q})^4=1-(\dfrac{m}{n})^2=(1-\dfrac{m}{n})(1+\dfrac{m}{n})$

$=(\dfrac{n-m}{n})(\dfrac{n+m}{n})=\dfrac{(n-m)(n+m)}{n^2}$, here $n^2 \nmid (n-m)(n+m)$ since $q \neq 1$.

Let $d$ be a divisor of $n$, note that $d \nmid(n+m)$ and $d \nmid (n-m)$, Using the fact: $\gcd(m,n)=1$

Now this implies $(n-m)(n+m)=p^4, n^2 =q^4$, which isn't possible.(Recalling Area of right angled triangle can't be a square, it doesn't exist as proven by Fermat)

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Denote $x=\dfrac{b}{a},y=\dfrac{d}{c},a,b,c,d>0,$ then $(\dfrac{b}{a})^4-(\dfrac{d}{c})^2=1,$ hence $a^4=c^2,c=a^2,b^4-d^2=a^4,$ this equation has no positive integer solution, see Solving $x^4-y^4=z^2$

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  • $\begingroup$ I didn't understand the 'hence' part there? Why $a^2=c^2,b^4-d^2=a^4$? $\endgroup$ – Inceptio May 21 '13 at 5:43
  • $\begingroup$ @Inceptio since $(a,b)=(c,d)=1$ and a rational number plus $1$ will not change the denominator,so $a^4=c^2$. $\endgroup$ – Next May 21 '13 at 7:09

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