We use this: If integers $ a,b,c,d $ satisfy $ ab=cd $, then there exist integers $ x,y,z,w $ that satisfy $ a=xy, b=zw, c=yz, d=wx $.

I think that: $(a^2+b^2)^2+a^2b^2=(m^2+n^2)^2 It's easy to show this equation hasn't ....

$a^2+b^2=m^2-n^2$ $ab=2mn$ $(a+b)^2=m^2-n^2+4mn$ $a+b=\sqrt{m^2-n^2+4mn}$ Let $a$ and $b$ are roots of this equation: $x^2-x\sqrt{m^2-n^2+4mn}+2mn=0$ $\Delta=m^2-n^2+4mn-8mn=m^2-n^2-4mn$ $a=\frac{\sqrt{m^2-n^2+4mn}+\sqrt{m^2-n^2-4mn}}{2}$ $=\sqrt{\frac{m^2-n^2+4mn+m^2-n^2-4mn+\sqrt{(m^2-n^2)^2-(4mn)^2}}{4}}$ $=\sqrt{\frac{2(m^2-n^2)+\sqrt{(m^2-n^2)^2-(4mn)^2}}{4}}$ $=\sqrt{\frac{2(a^2+b^2)+\sqrt{(a^2+b^2)^2-(2ab)^2}}{4}}$ $=\sqrt{\frac{2(a^2+b^2)+\sqrt{(a^2+b^2-2ab)(a^2+b^2+2ab)}}{4}}$ $=\sqrt{\frac{2(a^2+b^2)+\sqrt{(a+b)^2(a-b)^2}}{4}}$ $=\sqrt{\frac{2(a^2+b^2)+(a+b)(a-b)}{4}}$ $=\sqrt{\frac{2(a^2+b^2)+a^2-b^2}{4}}$ $=\sqrt{\frac{3a^2-b^2}{4}}$ $4a^2=3a^2-b^2$ $-(a^2+b^2)=a^2+b^2=0$ $a=b=0$ We can see that $m=n=0$

Can anyone show me how to write square root?

\sqrt{expression} Here's an example: $\sqrt{3}$ is written in latex as \sqrt{3}

math90 wrote: \sqrt{expression} Here's an example: $\sqrt{3}$ is written in latex as \sqrt{3} Thanks!

quaNGuyen wrote: $=\sqrt{\frac{m^2-n^2+4mn+m^2-n^2-4mn+\sqrt{(m^2-n^2)^2-(4mn)^2}}{4}}$ You miss the $2$ before $\sqrt{(m^2-n^2)^2-(4mn)^2}$. Thus your solution is not valid, because by putting that $2$ into its place gives the trivial identity $a=a$.

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