For any positive integers $a$ and $b$, the number $(36a+b)(a+36b)$ can never be a power of $2$. APMO 1998:

Show that for any positive integers $a$ and $b$, the number
  $(36a+b)(a+36b)$ can never be a power of $2$.

The solution I've read substitutes $a=2^Ap,b=2^Bq$ where $p$ and $q$ are odd positive integers and derives a contradiction.But is my method correct?
EDIT: The method below has some incompleteness.Please refer to Andre Nicolas' answer below to know what else I should have mentioned.In brief,I forgot to think(and mention in my solution) why the descent should occur in the first place.
My method: $$(36a+b)(a+36b)=2^m$$
This implies that $$36a+b=2^K$$,$$36b+a=2^L$$ Clearly,$a$ and $b$ both must be even.Let $a=2a'$ and $b=2b'$. Substituting these in the two equations,we obtain similar equations $$36a'+b'=2^{K-1}$$ and $$36b'+a'=2^{L-1}$$ By the method of infinite descent,can we now conclude that there are no solutions to the  above two equations?
P.S:The above assumes that  $a$ and $b$ aren't both $2$.
 A: Nice proof!  This kind of proof, an "infinite" construction based on an assumption that is false (and that we want to prove false), can be misunderstood. So it is useful to make the steps fully explicit. We give three ways of doing so, the last of which is descent.
1: Fermat's Method of Infinite Descent is, essentially, induction, in the form of what is often called the least number principle: Any non-empty set of non-negative integers has a smallest element.
Suppose then that there exists a a non-negative integer $u$ for which there are positive integers $x$ and $y$, such that $(36x+y)(x+36y)=2^u$.  Then there is a least such integer. Call it $k$. Let $a$ and $b$ be positive integers such that 
$$(36a+b)(a+36b)=2^k.\tag{1}$$
In (1), the product of two integers $36a+b$ and $a+36b$ is a power of $2$. Thus each of $36a+b$ and $a+36b$ is a power of $2$. Since each of $36a+b$ and $a+36b$ is $\ge 37$, each is is a power of $2$ that is $\ge 2^6$. 
In particular, $36a+b$ and $a+36b$ are each even. It follows that $a$ and $b$ are even. 
Let $a=2a_1$ and $b=2b_1$. Then $36a+b=2(36a_1+b_1)$ and $a+36b=2(a_1+36b_1)$. It follows that $(36a_1+b_1)(a_1+36b_1)=2^{k-2}$. It is clear that $k-2$ is non-negative. (Indeed $(k-2)\ge 12$.) This contradicts the supposed minimality  of $k$.

2: One can also rephrase the proof as a conventional induction proof. We want to show that for all non-negative integers $n$, there are no positive integers $a$ and $b$ such that $(36a+b)(a+36b)=2^n$. The result is trivial to verify for $n=0$ and $n=1$. Next we prove the induction step, that if the result is true at $n=k-2$, then the result is true at $n=k$. 
Suppose to the contrary that $(36a+b)(a+36b)=2^k$. Arguing exactly like above, we show that there exist $a_1$ and $b_1$ such that $(36a_1+b_1)(a_1+36b_1)=2^{k-2}$, contradicting the induction assumption. 

3: Infinite Descent uses exactly the same arithmetical observations, but the packaging is different. Let us suppose that there are positive integers $a_0,b_0$ and a non-negative $k_0$ such that  $(36a_0+b_0)(a_0+36b_0)=2^{k_0}$. 
By the argument described in detail above,  there exist positive integers $a_1,b_1$ and a non-negative $k_1\lt k_0$ such that $(36a_1+b_1)(a_1+36b_1)=2^{k_1}$. 
But then there exist positive integers $a_2,b_2$ and a non-negative $k_2\lt k_1$ such that $(36a_2+b_2)(a_2+36b_2)=2^{k_2}$. And so on, forever.
Thus there exists an infinite descending sequence $k_0\gt k_1\gt k_2\gt \cdots$ of non-negative integers. But there cannot be such a sequence, so we have reached a contradiction.  
A: that is not a hard question !
$36a^2+36b^2+(36^2+1)ab=36((a+b)^2-2ab)+(36^2+1)ab=36(a+b)^2+(36\cdot34+1)ab=2^n$
$\Longrightarrow$$36(2^Ap+2^Bq)^2+(36\cdot34+1)2^{A+B}pq=2^n$
$\Longrightarrow$$36(p+2^{B-A}q)(q+2^{A-B}p)+(36\cdot34+1)pq=2^m$
$\Longrightarrow$$36\cdot2^{B-A}q^2+36\cdot2^{A-B}p^2+(72+36\cdot34+1)pq=2^m$
$\Longrightarrow$$36\cdot(q^2+p^2)+(72+36\cdot34+1)pq=2^m$
a contradiction holds !
is it helpful ?
A: Unfortunately, the infinite descent doesn't work because the one of RHSs will eventually cease (or even initially fail) to be even, as 1 is also a power of two. You need to consider what happens when you reach the cases $36a^*+b^*=2^0$ and/or $36b^*+a^*=2^0$ because then the reasoning $a^*$ and $b^*$ are even breaks down.
Which suggests the original solution which extracts all the powers of two at once might be necessary.
