Posted, more exactly: yesterday!

Let a_1, b_1 be positive integers satisfying the required condition. Then a_1=4a_2 and b_1=4b_2 for some a_2 and b_2 as a_1 and b_1 divisible by 4. So smaller a_2 and b_2 also satisfies the condition. But there are not infintely many positive integers less than a_1 and b_1, contradiction.

We see that d=gcd (a,b) is a power of 2. Let a=dA, b=dB. Cancelling both sides by d and observing, we get A and B are both even, contradiction.

Another approach - the system $ 36a + b = 2^s, 36b + a = 2^t$ has integer solutions iff $ 36^2 - 1 | 36 \cdot 2^t - 2^s, 362^s - 2^t$ (just solve the system explicitly). $ 36^2 - 1 = 35 \cdot 37$. $ 35, 37$ are coprime. $ 35 | 36 \cdot 2^t - 2^s, 36 \cdot 2^s - 2^t \leftrightarrow 2^{s - t} \equiv 1 \pmod {35}$ $ 37 | 36 \cdot 2^t - 2^s, 36 \cdot 2^s - 2^t \leftrightarrow 2^{s - t} \equiv - 1 \pmod {37}$ $ 2^{s - t} \equiv 1 \pmod {35} \implies s \equiv t \pmod {12}$ (because the order of $ 2$ is $ 12$) $ 2^{s - t} \equiv - 1 \pmod {37} \implies s \equiv t \pmod {18}$ but $ s \neq t \pmod {36}$ (because the order of $ 2$ mod $ 37$ is $ 36$, and $ 2$ is not a quadratic residue) Yet, $ 12,18 |a \implies lcm(12,18) = 36 |a$, a contradiction. [EDIT: Now when I think of it, it reminds me of some IMO problem. See this - http://projectpen.files.wordpress.com/2008/12/pen-16sol.pdf ]

shobber wrote: Show that for any positive integers $ a$ and $ b$, $ (36a + b)(a + 36b)$ cannot be a power of $ 2$. Assume that there exists a pair. Then there must a exist a pair such that $ (a,b) = 1$. Then $ 36a + b = 2^n, 36b + a = 2^m$. So $ n,m > 5$. Thus $ 2^5 \mid (36a+b)+(36b+a) = 37(a+b) \iff 2^5 \mid a+b$. Thus $ a \equiv b \equiv 1 \bmod 2$. And thus $ 36a + b \equiv 1 \bmod 2$. Contradiction..

let $ (a+36b)(36a+b)=2^z$. so $ 36a+b=2^x$ and $ a+36b=2^{x+y}$. (assume that a+36b is not less thsn 36a+b). So $ 36a.2^y+b.2^y=2^{x+y}$ . Hence $ y<6$. but $ 37|2^x+2^{x+y}$.absurd.

Assume there is a solution. Take $a < b$ and the smallest possible $a$. Now $(36a + b)$ and $(a + 36b)$ must each be powers of $2$. Hence $4|b$ and $4|a$. So $a/2$ and $b/2$ is a smaller solution with $a/2 < b/2$. Contradiction.

First $a$ and $b$ must be distint so WLOG $a>b$ so $36b+a$ divides $36a+b$ hence $36b+a$ divides $a-b$ (impossible)

WLOG assume that $v_2(b)\ge v_2(a)$. Then we can divide by by $2^{2v_2(a)}$ to get $(36m+2^km)(2^k\cdot 36n+m)$ where $m$ and $n$ are odd positive integers. Since $2^k\cdot 36n+m$ is odd and not equal to $1$, then $(36a+b)(a+36b)$ cannot be a power of $2$.

subham1729 wrote: First $a$ and $b$ must be distint so WLOG $a>b$ so $36b+a$ divides $36a+b$ hence $36b+a$ divides $a-b$ (impossible) How do you conclude that 36b+a must divide a-b from that fact that 36b+a divides 36a+b?

Would this solution be acceptable?

It looks fine, just make sure the grader knows that you know that $2^k\geq 2^2$

mathlete3141592 wrote: subham1729 wrote: First $a$ and $b$ must be distint so WLOG $a>b$ so $36b+a$ divides $36a+b$ hence $36b+a$ divides $a-b$ (impossible) How do you conclude that 36b+a must divide a-b from that fact that 36b+a divides 36a+b? Does anybody see why this is true?

i think it should be $35a-35b$ by Euclid's Algorithm, I can't get any farther

rafayaashary1 wrote: i think it should be $35a-35b$ by Euclid's Algorithm, I can't get any farther Then why can't 36b+a divide 35a-35b? I can't understand subham's solution.

mathlete3141592 wrote: subham1729 wrote: First $a$ and $b$ must be distint so WLOG $a>b$ so $36b+a$ divides $36a+b$ hence $36b+a$ divides $a-b$ (impossible) How do you conclude that 36b+a must divide a-b from that fact that 36b+a divides 36a+b? Remember we are assuming that $36b+a$ is a power of two. Then $(36b+a,35)=1$ so $36b+a|35a-35b$ implies $36b+a|a-b$...

$(36a+b)(36b+a)=2^{m}$. Let $36a+b=2^{t}$ $36b+a=2^{g}$ Therefore $37a+37b=2^{t}+2^{g}=2^{\min{(g,t)}}(2^{|g-t|}+1)$. Therefore $2^{\min{(g,t)}} | a+b$. Contradiction

Solution WLOG suppose $ b \le a $ Then clearly $ (36b +a ) | (36a +b) $ Therefore $ gcd({ 36a + b , 36b + a }) = gcd({ 36a +b , 35(a-b) }) = 36b + a$ $ \implies $ $ (36b + a) | 35(a-b)$ Thus either $ (36b +a) | 35$ ........$ (1) $ OR $ (36b + a ) | (a-b) $..........$(2) $ $1$ is absurd as $ a,b \in Z+ $ and $2$ is clearly impossible.

Just write $a=2^{\alpha}a',b=2^{\beta}b'$ to get $2^{\alpha +2} \cdot 9a' +2^{\beta}b'=2^m$, and $2^{\beta+2} \cdot 9b' +2^{\alpha}a'=2^n$- thus, as $a,b$ are positive integers, $\alpha+2=\beta, \beta+2=\alpha$ which is absurd.

I'm not quite sure about this, can someone or Point any flaws? APMO 1998 P2 wrote: Show that for any positive integers $a$ and $b$, $(36a+b)(a+36b)$ cannot be a power of $2$. Solution: Since, $(36a+b)(a+36b)$ is symmetric. Hence, Suppose, $a \geq b$. Now suppose, there exists a $k$ $$(36a+ ~b)(a+36b)=2^k=2^{x+y}$$Suppose, $$\begin{cases} 36a+ ~b=2^x \\ 36b+ ~a=2^y \end{cases}$$ Hence, $a,b$ are even $\implies$ $a=2a_1$ and $b=2b_1$ for $a_1,b_1 \in \mathbb{Z}^+$ (#27) Thanks $$\begin{cases} 36a+b=2(36a_1+b_1)=2^x \implies 36a_1+b_1=\frac{2^x}{2} =2^{x-1}=2^{x_1} \in \mathbb{Z}^+ \\ 36b+a=2(36b_1+a_1)=2^y \implies 36b_1+a_1=\frac{2^y}{2} =2^{y-1}=2^{y_1} \in \mathbb{Z}^+ \end{cases}$$Hence, now, $$(36a_1+b_1)(36b_1+a_1)=2^{x_1+y_1}$$Repeating the Same Procedure, we find, $a_1=2a_2$ and $b_1=2b_2$ for $a_2,b_2 \in \mathbb{Z}^+$. We can continue This till $x_n+y_n$ doesn't become negative and Since, $a,b$ are positive integers, we get a Contradiction! $\qquad \blacksquare$ @Jupiter and agastya, Thanks

Redacted

Note that $x_1+y_1<x+y$ and thus going towards $-\infty$. Therefore, there will be no infinite descent because it will be stop when $x_n+y_n<0$. But you may lead the proof : (i) When $x+y$ is even : In this case once you will get $x_n+y_n=0$ which is absurd since $m,n $ are positive integers. (ii) When $x+y$ is odd : In this case once you will get $x_n+y_n=1$ which is again not possible since $m,n$ are positive integers

@pluto1728 I don't understand what are you telling. I think his solution is completely fine. Btw @alastor From the condition $$\begin{cases} 36a+ ~b=2^x \\ 36b+ ~a=2^y \end{cases}$$Isn't it obvious that both of $a,b$ are even. Thanks Jupiter.

@above Lol stupid me... @Jupiter...I meant there is decreasing sequence of positive integers right ($a,b $)? So FMID should work? @below Oh! I kind of get it... Thanks! Mr.Chagol wrote: ^^ try solving this with geo Whut? How ?

AlastorMoody wrote: @above Lol stupid me... @Jupiter...I meant there is decreasing sequence of positive integers right ($a,b $)? So FMID should work? The argument that you used to proof that $a$ and $b$ are even used the fact that $x+y>0$. However, the sequence $\{x_i+y_i\}_{i=1}^{\infty}$ is strictly decreasing by a gap of $2$. Thus, once $x_i+y_i$ becomes negative (for some $i$), you lose the power to prove that $a_i\equiv b_i\equiv 0\pmod{2}$

^^ try solving this with geo @below thanks for reminding

Apmo 1998

https://artofproblemsolving.com/community/c2113h1042655

$\spadesuit \ \textbf{\textrm{A different solution without infinte decent:}}$ Let $a=2^c \cdot p, b=2^d \cdot q$ where $p,q$ are odd numbers and $c,d$ are non-negative integers, WLOG, assume that $c\ge d$, \[(36a+b)(a+36b)=2^d(36\cdot 2^{c-d} \cdot p+q)(36b+a)\]since $36\cdot 2^{c-d} \cdot p+q$ is a non-trivial odd positive integer, $(36a+b)(a+36b)$ cannot be a power of $2$. $\quad \blacksquare$

$36a+b$ and $36b+a$ are both powers of $2$ we can assume WLOG that : $a+36b|b+36a$ $\implies a+36b|35(a-b)$ but $a+36b$ is a power of $2$ so $a+36b|a-b$ that gives $a=b$ and that also is a contradiction