Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

I was trying do a problem I was stuck , the only remaining to prove was that $2^a+2^b=37k$ where $a(>1),b(>1)$ and $k$ are integers is not possible.

share|improve this question
9  
$2^{18} + 1 = 37\cdot 7085$. Multiply by any power of $2$. –  Daniel Fischer Aug 12 '13 at 12:44
1  
edit a and b must be greater than 1 –  maths lover Aug 12 '13 at 12:45
    
@DanielFischer thnx.. –  maths lover Aug 12 '13 at 12:46
    
Ah, well, maybe we can help you solve the problem in some other way? –  Daniel Fischer Aug 12 '13 at 12:46

1 Answer 1

Using $2^{18} + 1 = 37 \cdot 7085$ you can simply take $a = 20$ and $b = 2$ for example and you get $$2^{20}+2^2 = 37\cdot(7085\cdot 4)$$

Any other multiple of two also works as a counter example. The assertion is wrong.

share|improve this answer

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

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