# $a, b \in\Bbb N$, find all solutions to $2^a = b^2 - 5$ and prove there are no more solutions?

I am currently studying discrete mathematics at uni (in my computer science degree). We have an assignment due tomorrow, and i have been able to do most of it, but one question eludes me. I spoke to a tutor today about it, and he said that last year they asked the same question in the assignment, and only 4 people got it right in the entire course.

The question is this: given the possible remainders that a perfect square leaves when divided by $3, 4$ and $8$......., $a, b$ are in the natural numbers, find all solutions to $2^a = b^2 - 5$ and prove there are no more solutions than the ones you have found?

I have been trying all day to solve this, but i am still no closer.

I am guessing it has something to do with using a different modulus (one of the tutors hinted this), but i cant figure it out.

If someone wouldn't mind perhaps pointing me in the right direction, i would be hugely grateful.

Thanks Corey B :)

Hint What is $b^2 \pmod{8}$?..

This Diophantine equation is well-known. It has exactly the solution $a=2$ and $b=3$ in nonnegative integers, see Theorem $1$ here. In fact, the Diophatine equation $$2^x+5^y=z^2$$ has the only solutions $(x,y,z)=(3,0,3),(2,1,3)$ in nonnegative integers.

• That was perfect! Thanks heaps :) Apr 1 '15 at 12:39
• You are welcome. Apr 1 '15 at 12:46