*Originally posted by David113*

**Prove:
**

the only integer solutions of ab+5cd=3, ad=bc are:

a=1, b=3, c=d=0

a=-1, b=-3, c=d=0

a=3, b=1, c=d=0

a=-3, b=-1, c=d=0

I guess I'll have a go at this. I'll consider 4 different cases that collectively exhaust possibilities.

First, consider c=0 and d is nonzero. For this case we have ab=3; ad=0; d nonzero. This doesn't have any solutions because we require a=0 to satisfy the second equation, but that in turn would preclude any integer solutions to the first equation. No integer solutions for this case.

Second, consider c is nonzero, d=0. For this case we have ab=3; bc=0; c nonzero. Like the first case, this doesn't have solutions because we require b=0 to satisfy the second, and then we don't have any way to satisfy the first. No integer solutions for this case.

Third, consider c=d=0. Then we have ab=3, which yields the four integer solutions you listed and no others.

So we are left with considering cases where c and d are both nonzero integers. I will try to show that this, like the first two cases above, has no solutions. We have

(1) ab=3-5cd and (2) ad=bc; c and d nonzero.

Substituting (2) into (1) we get the following equations, which both must hold:

(3) (cb^2)/d = 3-5cd and (4) (da^2)/c = 3-5cd.

These respectively yield

(5) b^2 = 3(d/c) - 5d^2 and (6) a^2 = 3(c/d) - 5c^2

Adding (5) and (6) together, we get

(7) a^2 + b^2 = 3[(c/d) + (d/c)] - 5(c^2 + d^2), which can be rearranged to

(8) a^2 + b^2 = [3/(cd) - 5](c^2 + d^2)

I think it's clear that this has no integer solutions because the LHS would have to be greater than or equal to zero, whereas the RHS can only be strictly negative for c and d nonzero integers.

So I think the above cases exhaust possibilities, and they yield only the four solutions you listed.