<s>[INST] $x^2-y^2=196$, can we find the value of $x^2+y^2$? $x$ and $y$ are positive integers. 
If $x^2-y^2=196$, can we know what the value of $x^2+y^2$ is?
Can anyone explain this to me? Thanks in advance.
 [/INST] Rewrite the equation to get $x^2=y^2+196$. Now write $x=y+k$, then $$x^2=(y+k)^2=y^2+2yk+k^2=y^2+196$$
Therefore $2yk+k^2=196$
Now note that $k \mid 196$ and $k$ must be even.
This leaves the possibilities $2,4,14,28,98,196$. 
$k=2$ gives $y=48$.
$k=4$ gives $y=22.5$, which is not an integer. 
$k=14$ gives $y=0$, but that is not positive. 
Similiar, for larger $k$, it won't be positive.
So $y=48$, therefore $$x^2+y^2=(x^2-y^2)+2y^2=196+2\cdot48^2=4804$$
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