How many pairs of integers $(x,y)$ satisfy the equation $x^{2}-y^{2}=66$

Question

I came up with the expression $(x+y)(x-y)=66=3.2.11$

Now how to consider different cases to calculate the number of pairs?

Answer 1

You know that both $x+y$ will divide $66$ and $x-y$ will divide $66$. There are only a couple of options for their values:

$$33\cdot 2\\ 22\cdot 3\\ 11\cdot 6$$

You can check each of these three options to see whether you get an integer solution. For example, if $x+y=33$ and $x-y=2$, then $2x=35$ means $x=\frac{35}{2}$ so $x, y$ are not integers.

Answer 2

Answer is simply $0$ pair of integers.

Since, either $x^2-y^2$ will be divisible by 4 or be an odd no. In this case it is neither ($x^2-y^2$ is divisible by 2).

$(x+y)$ and $(x-y)$ will be of same parity.

Answer 3

We are looking for solutions to $$(x-y)(x+y)=66$$ Note that $$2x=(x+y)+(x-y)$$ is even.

That is $(x+y)$ and $(x-y)$ are both even or both odd.

Thus the product is either a multiple of 4 or an odd integer.

The product is 66 which is neither odd nor a multiple of $4$

Thus there are no solutions.