<s>[INST] When is $(a^2+b)(b^2+a)$ a power of $2$? When is $(a^2+b)(b^2+a)$ a power of $2$? 
($a$, $b$ positive integers)
I tried some values on the computer and it seems the only solutions are $a=1$, $b=1$. But I am not sure how to prove it?
Thanks for any help.
 [/INST] If $a$ has lower power of $2$ than $b$:
  $a+b^2$ will be an odd multiple of a power of $2$ and hence won't be a power of $2$
Similarly if $b$ has lower power of $2$ than $a$
Therefore $a,b$ must have the same power $p$ of $2$
Let $a = 2^p x$ where $x$ is odd
Let $b = 2^p y$ where $y$ is odd
Then $(a^2+b)(b^2+a) = 2^{2p} ( 2^p x^2 + y ) ( 2^p y^2 + x )$
Thus $2^p x^2$ is odd, and hence $p = 0$, and so $a,b$ are both odd
If $a = b$:
  $(a^2+b)(b^2+a) = a^2 (a+1)^2$
  Thus $a = 1$ otherwise $a^2$ is not a power of $2$
If $a \ne b$:
  $(a-b)(a+b-1) = 2^q z$ where $q$ is positive and $z$ is odd
  $a = 2^q + b$
  $(a^2+b)(b^2+a) = ( 2^{2q} + 2^{q+1} b + b^2 ) ( 2^q + b^2 + b )$
  Thus $2 | b$ otherwise $2^{2q} + 2^{q+1} b + b^2$ will not be a power of $2$
  Contradiction
Therefore $(a,b) = (1,1)$
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