Re: a square?

On May 4, 4:43 pm, Kent Holing <K...@xxxxxxxxxxx> wrote:
For m integers, when is (-4m+17)(m^2-4) squares?
Kent Holing
Norway

For a and b integers, a*b is a square for any of the following:
a = b
a = 0
b = 0

Therefore in your case:

-4m+17 = m^2-4 => m^2+4m-21=0 => m = {-7, 3}.

-4m + 17 = 0 => m = 17/4 which is not an acceptable solution in Z.

m^2-4 = 0 => m = {-2, 2}

Hence the solution is m = {-7, -2, 2, 3}.

HTH
Antonio

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