<s>[INST] $5(a^2+b^2)$ covers all numbers $=a_2^2+a_2^2=b_1^2+b_2^2$? I start by noting that 4a*2b=2a*4b
I write 4a*2b as $((2a+b)+(2a-b))*((2a+b)-(2a-b))  = (2a+b)^2-(2a-b)^2$
I follow a similar principle for 2a*4b which I write as $(a+2b)^2-(a-2b)^2$
to arrive at 
$(2a+b)^2-(2a-b)^2= (a+2b)^2-(a-2b)^2$ or
$(2a+b)^2+(a-2b)^2= (a+2b)^2+(2a-b)^2=5(a^2+b^2)$
 The expression $5(a^2+b^2)$ thus gives numbers which can be represented as a sum of two squares in two different ways, and I observe that if a=b, a=2b (or b=2a) we have simpler relations such as (a,b)=(1,1) which gives us $5*2=3^ 2+1 $ 
I observe that we cover $50=7^2+1^2=5^5+5^5$, $65=8^2+1^1=7^2+4^2$ etc.
I now wonder does this simple formula cover all such numbers? 
This doesn’t necessarily follow from my simple calculation. 
 [/INST] $$ 13 (a^2 + b^2) = (3a+2b)^2 + (2a-3b)^2 =  (3a-2b)^2 + (2a+3b)^2$$
$$ 17 (a^2 + b^2) = (4a+b)^2 + (a-4b)^2 =  (4a-b)^2 + (a+4b)^2$$
$$ 29 (a^2 + b^2) = (5a+2b)^2 + (2a-5b)^2 =  (5a-2b)^2 + (2a+5b)^2$$
$$ 37 (a^2 + b^2) = (6a+b)^2 + (a-6b)^2 =  (6a-b)^2 + (a+6b)^2$$
$$ 41 (a^2 + b^2) = (5a+4b)^2 + (4a-5b)^2 =  (5a-4b)^2 + (4a+5b)^2$$
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