Hey ! Totally contradicting the Fermat's 4n+1 theorem ,isn't it??

Hey, in the problem you take all such $4n+1$ for which $n$ is a square and you want to express it as $a^2+b^2$ where both a,b>1, so I don't think it contradicts the Fermat's theorem on sums of two squares.

Lemma: If $n=x^2+y^2=z^2+t^2$ and $x\ne z,t$ then n is a composite number Suppose $2|x-z$ and $2|y-t$ $\implies n=\left(\frac{x+z}{2} \right)^2+\left(\frac{x-z}{2} \right)^2+\left(\frac{y+t}{2} \right)^2+\left(\frac{y-t}{2} \right)^2=p^2+q^2+r^2+s^2$ Let $a=(p,r)$ then $p=ab,r=ac$ Because $pq=rs$ so $abq=acs\implies b|cs$ Because $(b,c)=1$ so $b|s\implies s=bd\implies q=cd$ $\implies n=(ab)^2+(cd)^2+(ac)^2+(bd)^2=(a^2+d^2)(b^2+c^2) \implies n$ is a composite number