APMO

Suppose that $(36a+b)(a+36b)$ is a power of $2$ for some positive integers $a$ and $b$. Write $36a+b=2^m=r$ and $a+36b=2^n=s$. Then $$36r-s=35\times 37a,\text{and}36s-r=35\times 37b.$$

Hence $$\frac{1}{36}<\frac{r}{s}=2^{m-n}<36,\text{or}-6<m-n<6.$$

Furthermore, $$4^n(4^{m-n}-1)=r^2-s^2=35\times 37(a^2-b^2).$$

Thus $$4^{m-n}\equiv 1(\mathrm{mod}37).$$

Observe that the $9$th power of $4$ is the smallest power of $4$ that is congruent to $1$ modulo $37$. Thus $9\mid (m-n)$. Also note that $m\ne n$. Hence $|m-n|\ge 9$, which is not possible because $|m-n|<6$.