## eliassaab 3 years ago Let m and n be two positive integers. Show that (36 m+ n)(m+36 n) cannot be a power of 2.

1. sauravshakya

LET (36m+n)=2^x AND (m+36n)=2^y THEN, (36m+n)(m+36n)=2^(x+y) AlSO, x and y must me both integers

2. sauravshakya

Now, n=2^x - 36m THEN, m+36n=2^y m+36(2^x-36m)=2^y 36 * 2^x -1295m=2^y

3. sauravshakya

Now, I think I have to prove that y is never a integer

4. sauravshakya

2^y=36*2^x-1295m 2^y=2^x(36-1295m/2^x) Now, 2^y must be positive so, (36-1295m/2^x) must be positive... Now let 2^z=36-1295m/2^x HERE z also must be an positive integer..... So, 2^z can be 4,8,16,32 NOW,

5. sauravshakya

36-1295m/2^x= 4, 8, 16, 32 1295m/2^x=32, 28 ,16 ,4 2^x=1295m/32 , 1295m/28 , 1295m/16 , 1295m/4

6. sauravshakya

Now, 2^x=1295m/32 , 1295m/28 , 1295m/16 , 1295m/4 2^x=40.47 m , 46.25m , 80.94m ,323.75m Thus, for no positive integer value of m we will get x a integer value......

7. sauravshakya

Hence, (36 m+ n)(m+36 n) cannot be a power of 2.

8. sauravshakya

I have jumped some step......... I hope I made it clear

9. sauravshakya

@eliassaab

10. sauravshakya

@ganeshie8

11. mukushla

for $$m=n$$ the statement is true suppose $$m\ge n$$ in order for $$(36 m+ n)(m+36 n)$$ to be a power of $$2$$ $36m+n=2^a$$m+36n=2^b$$a> b$$\frac{36m+n}{m+36n}=2^{a-b}=2^c \ \ \ c\ge 1$$m+36n | 36m+n$$m+36n|36(m+36n)-(36m+n)=1295n$since $$\gcd(m+36n \ , \ n)=1$$ so $$m+36n$$ is a divisor of $$1295$$ so $$(36 m+ n)(m+36 n)$$ is not a power of 2

12. experimentX

looks like gcd is very important ... i never liked it.

13. eliassaab

Here is my proof. Suppose not, let m and n be the smallest such that this product is a power of 2. It is easy to see that m and n are divisible by 2. So m = 2 M and n = 2 N. Since ( 36 (2M) + 2 N)(36(2N)+2 M) is a power of 2,then ( 36 ( M) + N)(36( N+ M) is a power of 2 (contradiction).

14. mukushla

Short and Neat :)