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Prove that for any integers $a, b, c, d$ number $(a^2 + b^2)(c^2 + d^2)$ is a sum of two squares of integer.

In fact I have no idea how to do this and I'll appreciate any tips or the solution.

tdudzik
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\begin{align} (a^2+b^2)(c^2+d^2)& =a^2c^2+a^2d^2+b^2c^2+b^2d^2 \\[10pt] & =a^2c^2+2abcd+b^2d^2+a^2d^2-2abcd+b^2c^2 \\[10pt] &=(ac+bd)^2+(ad-bc)^2 \end{align}

Michael Hardy
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Roman83
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There is a nice way using complex numbers:

Let $z=a+bi, w=c+di$. Then $|z|^2=a^2+b^2$ and $|w|^2=c^2+d^2$, so $$|z|^2|w|^2 = (a^2+b^2)(c^2+d^2).$$ On the other hand, we have $zw=(a+bi)(c+di)$. Now calculate $|zw|^2$ by squaring both sides and use the fact that $|zw|^2=|z|^2|w|^2$.

Théophile
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  • Can this idea be extended to show that if $a^2,b^2$ and $c^2,d^2$ are distinct then we can write $(a^2+b^2)(c^2+d^2)=x^2+y^2=p^2+q^2$? – MHW Nov 09 '16 at 22:00
  • @user332597 I believe so, provided that $a,b,c,d$ are non-zero. From the formula, we have that $(a^2+b^2)(c^2+d^2)=(ac-bd)^2+(ad+bc)^2$. If we switch $c$ and $d$ we get $(ad-bc)^2+(ac+bd)^2$, and now suppose that these two squares are the same as before. Then either $(ac-bd)^2=(ad-bc)^2$ or $(ac-bd)^2=(ac+bd)^2$, and both cases end in contradiction (either $a^2=b^2$ or one of the terms is $0$). – Théophile Nov 11 '16 at 04:18
  • @user332597 See also this MathWorld article on the number of ways to express an integer as the sum of squares. – Théophile Nov 11 '16 at 04:19
  • This is great, thank you very much. – MHW Nov 11 '16 at 19:43
  • @user332597 You're welcome! – Théophile Nov 11 '16 at 20:23