Without the geometry, the most simple way is this. Let **b**^2 be the square of the smallest possible natural **b**, such that it is twice as big as another square of a natural, **a**^2!

2***a**^2=**b**^2* (1)*

2***a**^2=4*(**b**/2)^2 *(2)*

**a**^2=2*(**b**/2)^2 *(3)*

Now, if the **b**/2 is a natural, even a smaller pair of naturals satisfies the initial condition. And **b** must indeed be even, since **b**^2 is even.

*(1) *is a direct consequence, or another form of: **a**/**b**=SQRT(2).

*DISCLAMER:*

*This one has been known since long ago, nothing new here in this post. Just a small clarification of the previous, much more ambitious post.*

*Done with that.*

Advertisements

Actually I think using mod 3 might be even simpler. Squares are 0 or 1 mod 3, so b^2 is 0 or 1 and 2a^2 is 0 or 2. Hence a and b divide by 3, contradiction.

This is faster than the mod 2 proof above because you don’t have to rewrite the equation after you discover b is even.