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.

  • $\begingroup$ If $a=b$, then both must be $1$. $\endgroup$ – Indrayudh Roy Jun 23 '14 at 15:33
  • 1
    $\begingroup$ By the way, where did this question come from? Quite a strange one to me. $\endgroup$ – user21820 Jun 23 '14 at 16:14
  • $\begingroup$ I thought of this question myself, just experimenting :) $\endgroup$ – yoyostein Jun 23 '14 at 16:52

The product $(a^2+b)(a+b^2)$ is a power of two if and only if both factors are: $$\begin{array}{rcl} a^2+b&=&2^r\\ a+b^2&=&2^s \end{array}$$

If $a=b$, then the equation $a^2+a=a(a+1)=2^r$ has only one solution, namely, $a=1$. So we can WLOG assume that $a>b$. Thus, $r>s$.

Since $2^s=a^2+b\geq2^2+1$, we have that $s>1$.

Since $s>1$, $a$ and $b$ have the same parity. Substracting the equations and factoring we have: $$(a-b)(a+b-1)=2^s(2^{r-s}-1)$$ Since $a+b-1$ is odd, $2^s$ divides $a-b$, that is, $a=2^sk+b$, $k\geq1$. But $a=2^s-b^2<2^s$, a contradiction.

  • $\begingroup$ that was fast. thanks! $\endgroup$ – yoyostein Jun 23 '14 at 16:51
  • $\begingroup$ Are you interested in the equation $(a^2+b)(a+b^2)=p^n$ for $p$ prime? $\endgroup$ – ajotatxe Jun 24 '14 at 5:05

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 $2$


Therefore $(a,b) = (1,1)$


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.