When is 2m^4 + 4m^3 + 2m^2 + 1 a Square?

F(m) = 2m^4 + 4m^3 + 2m^2 + 1 is a perfect square for m = 1 and m = 3 (at which it takes the values 3^2 and 17^2, respectively). Are there other positive integer values of m which make F(m) a perfect square? If not, can you prove your assertion?

Any suggestions on tackling this problem?

Joseph L. Pe

The only integer solutions for F(m) are 1, 9, and 289, corresponding respectively to m = {0,-1}, {1,-2}, and {3,-4}. So m=1 and m=3 are indeed the only positive integer values.

A complete proof can be found at:

KIND OF SPOILER - Method rather than Solution

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1) No. 2) Yes.

Curve sketching would probably be your best (and easiest) bet. A quartic like this will have 2 minimums, 1 maximum and two points of inflexion.

Exchanging f(m) for y and m for x:

x-intercepts [Use y = 0] y-intercepts [Use x = 0] local maxima [Use dy/dx = 0, look for sign change + to -] local minima [Use dy/dx = 0, look for sign change - to +] points of inflection [Use d2y/dx2 = 0, and the sign of d2y/dx2 changes]

You can then sketch y=x^2 over the top of this curve to deduce that the two lines will cross only twice for for positive x.

This only proves that the equation 2x^4 + 4x^3 + 2x^2 + 1 = x^2 has only 2 positive real roots.

The proof looks quite complex. Is there perhaps a simpler and more elegant proof?

The question already specified that a) we were looking for positive integer solutions and b) that at least 2 existed, namely m=1 and m=3. Thus sketching the curve would show that there were only 2 positive real roots and further we can assume that these must be those already given. The question was to prove there were no others and the sketch does that.

The original problem can't be reduced to finding the intersections of the two curves in 2x^4 + 4x^3 + 2x^2 + 1 = x^2. As you already know, m = 1 and m = 3 are solutions, but they don't satisfy the above equation. (Try plugging in x = 1 or x = 3.) The problem is really to find lattice points (x,y) (i.e., points with integer coordinates) of y = sqrt(2x^4 + 4x^3 + 2x^2 + 1 = x^2).

F(m) = 2m^4 + 4m^3 + 2m^2 + 1

for m = 1, F(m) = 3^2 and for m = 3 , F(m) = 17^2 what you tried to show was that F(m) = m^2 has 2 postive real roots, clearly m=1 and m=3 are *not* roots of F(m) = m^2.

in fact we can easily show that for all m > 0 , F(m) > m^2 i.e F(m) = m^2 has *no* positive real roots.