The sides of the three squares $ABCD,AEZH$ and $AUIK$ have integer lengths and $(AEZH)-(ABCD)=31$. If $BE=UE$, then calculate $(AUIK)$

Question

The sides of the three squares $ABCD,AEZH$ and $AUIK$ have integer lengths and $(AEZH)-(ABCD)=31$. If $BE=UE$, then calculate $(AUIK)$. Where $(A_1A_2A_3A_4)$ is the area of the quadrilateral $A_1A_2A_3A_4$

I tried solving it as follows:

I state that $BE=a$ and $x=AU$, then $(x-a)^2-(x-2a)^2=31$

$2ax-3a^2=31$

And I got stuck here. I believe the answer is 289, but I can't prove it. Could you please explain to me how to solve this question? This question is a multiple choice, the possible answers are 225, 256, 289, 300, 961

Answer 1

Knowing that the squares have integer lengths, $a,x$ are both positive integers.

Now $31 = 2ax - 3a^2 = a(2x - 3a)$. Hence both $a$ and $2x-3a$ are factors of $31$.

Fortunately $31$ is prime, so $a$ is either $1$ or $31$. This gives $2x - 3a = 31$ or $1$ respectively. What is the corresponding value(s) of $x$?