Can you show some of your attempts at these problems?
I presume that the solutions a = 0, b = anything, and a = anything, b = 0 are both already known to you, so we will go ahead and assume that neither a nor b are 0. As scounged suggested, expand both sides. After some amount of simplifying I get
This is a quadratic in a. Now take a look at the discriminant:
For an integer solution for a we must require that is a perfect square. Are there any integer b that do this? If you find solutions to b, use the quadratic equation in a,
to find if this gives an integer value for a.
-Dan
Factor all perfect squares from the RHS:
That means that we need a value of an integer b such that is a perfect square. If you don't see it, start picking values for b and see what you get. Note that, since b is squared all you need to do is check all the non-negative values for b. (And, of course once you do this, you still need to verify that these solutions for b give an integer value for a.)
Hint: There are only two values of b that make a perfect square. Can you prove this?
-Dan
Can't prove there are only two b values or that there is no integer a for either of these two b values?
First remove any perfect square factors from . So we require that must be a perfect square.
There are only two possibilities for b: b is even or it is odd.
If b is even then let b = 2n. This implies that . We need to be an even number divisible by 2 to make this a square number. But is odd. Thus b cannot be an even number.
So let b be an odd number: b = 2n + 1. Then . Now, 4 is a perfect square, so again I am going to factor that out of the problem. This leaves the statement that must be a square. By inspection n = 0 makes this a perfect square. So n = 0 implies b = 1. Further, the only other way that can be a perfect square is to let . This has the solution n = 1, which says that b = 3.
So we have only two possible solutions for an integer b: b = 1 or b = 3. Both of these give integer solutions for a.
-Dan