Any ideas?

https://drive.google.com/file/d/1W4iBrV_IhHRVgFsRddpkgTy9tAJfDhoP/view sorry I so bad English candlestick I can't help you a post by English. It's Vietnamese document. PAGE 17

Note that by taking ${\rm gcd}(a,b)$ away if necessary, we can assume $(a,b)=1$. Now let $p\mid a^2+b^2$ be a prime. Then $p\mid a^2+b^2\mid ac+bd$ implies $a^2c^2\equiv b^2d^2\pmod{p}$. Setting $b^2\equiv -a^2\pmod{p}$, we get $p\mid a^2(c^2+d^2)$. Now if $p\mid a$ then $p\mid a^2+b^2$ yields $p\mid b$, absurd as $(a,b)=1$. Hence $p\mid c^2+d^2$, yielding the conclusion.