Prove there is no integer solution to $a^2 - b^2 = 2$ by contradiction

Question

I have been trying this question for a long time, as a similar one will be in my test next week. I have done this:

$a^2 - b^2 = 2 $
$a^2 = 2 + b^2 $
$a = \sqrt2 + b$

As $\sqrt2$ is not an integer, and if b is an integer, adding them together will yield a non-integer which proves that there is no integer solution.

Welcome to MSE. Please learn to use MathJax to type the math in your questions. — jjagmath, Apr 19 '21 at 17:08
Welcome to Mathematics Stack Exchange. This could be proved considering remainders modulo $4$ — J. W. Tanner, Apr 19 '21 at 17:09
The two least possible difference of squares of integers are 1 and 3 , therefore no solutions @YvesDaoust — , Apr 19 '21 at 17:11

score 1 · Answer 1 · answered Apr 19 '21 at 17:11

1

$a-b$ and $a+b$ have the same parity, hence their product is odd or a multiple of $4$.

answered Apr 19 '21 at 17:11

score 0 · Answer 2 · answered Apr 19 '21 at 17:13

0

The two least difference of squares of integers are 1 and 3 , all the other difference of squares are greater than these ,

Therefore , no integer solutions

answered Apr 19 '21 at 17:13

You're correct (besides $0$), but can you prove it? – J. W. Tanner Apr 19 '21 at 17:15

Peter Szilas · Answer 3 · 2021-04-19T17:31:23.927

0

Can restrict to $a, b \ge 0$, since the expression is quadratic.

$(a-b) (a+b) =2;$

$2$ is a prime number, hence either

1)$a-b=1$, and $a+b=2$ or

2)$a-b=2$, and $a+b=1;$

No integer solutions for the above equations.

edited Apr 19 '21 at 17:31

answered Apr 19 '21 at 17:23

Peter Szilas

3 Answers3