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Let $m$ and $n$ be odd integers such that $m$ is the sum of two squares and $n$ is the sum of two squares. I am supposed to show that $2mn$ is sum of two odd squares.

First, since $m$ and $n$ are both sums of two squares, I showed that $m=4a+1$ for some integer $a$ and $n=4b+1$ for some integer $b$. Then, $$2mn=2(4a+1)(4b+1)=32ab+8a+8b+2.$$ It is correct to think along the lines of splitting $32ab+8a+8b+2$ into sums of two squares? If it is then would appreciate some help on this.

Bill Dubuque
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KHOOS
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    Hint: $(a^2+b^2)(c^2+d^2)=(ad-bc)^2+(ac+bd)^2$ shows you that $mn$ is the product of two squares. Now you have to handle the $2$ and also show that the squares involved are odd. – David Mar 22 '23 at 05:22
  • If $2(a^2+b^2)(c^2+d^2)=2(ad-bc)^2+2(ac+bd)^2$, then do I have to simplify the right hand side until they become sum of two odd squares? – KHOOS Mar 22 '23 at 06:00

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The Brahmagupta–Fibonacci identity shows that $mn$ is also a sum of two squares: $$mn=a^2+b^2$$ hence (applying Brahmagupta–Fibonacci identity again but in a very simple case: $2=1^2+1^2$) $$2mn=(a+b)^2+(a-b)^2.$$ Since $a^2+b^2$ is moreover odd, $a\pm b$ are odd.

Anne Bauval
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