# If ${ x }^{ 4 }+{ y }^{ 2 }=1$ then $x$ and $y$ can be both rational numbers?

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: $x,y\neq0$

• $\pm1,0$ and $0,\pm1$ are obvious solution – lab bhattacharjee May 20 '13 at 4:33
• Let's say between 0 and 1. – newzad May 20 '13 at 4:34
• – lab bhattacharjee May 20 '13 at 4:36
• @labbhattacharjee this question is different because I take square of a rational number then ask for the rationality of $\cos\alpha$ – newzad May 20 '13 at 4:39

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)

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 $x^4-y^4=z^2$

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