Question

Determine all positive integers $k \le 2000$ for which $x^4 + k$ can be factored into two distinct trinomial factors with integer coefficients

Answer 1

The answers are 4, 64, 324, and 1024. A proof is below.

Let the trinomial factors be
`ax^2 + bx + c` and
`dx^2 + ex + f`. Then, their product is
`(ad)x^4 + (ae+bd)x^3 + (fa+be+cd)x^2 + (bf+ce)x + cf`. We'll solve this problem by analyzing each of these coefficients.

Since the
`x^4` coefficient is
`ad` and all of the variables are integers, we have either
`a = d = 1` or
`a = d = -1`. Without loss of generality, let
`a=d=1`. (In the -1 case, we could simply multiply all the variables by -1 to find an analogous 1 case.)

So, we're back to
`x^2 + bx + c` and
`x^2 + ex + f`. We have that the
`x^3` coefficient is
`b+e`, and since that term is zero in this polynomial, we have
`b+e=0`. This means that
`e=-b`.

Our polynomials are now
`x^2 + bx + c` and
`x^2 - bx + f`. Let's consider the
`x` coefficient. It shows that we must have
`bf - bc = 0`, so either
`b = 0` or
`f - c = 0`.

Let's look at the case with
`b=0`. In this case, our polynomials become
`x^2 + c` and
`x^2 + f`. Then, we must have, for the
`x^2` coefficient to be zero, that
`c+f=0`, so
`c = -f`. But then, considering the constant term, we'd need
`-c^2 = k`. However, we are given that k is positive, and we know that
`-c^2` is negative, so this case doesn't work. Hence, we must have that
`f - c = 0`, so
`f = c`.

Our polynomials are thus
`x^2 + bx + c` and
`x^2 - bx + c`. Hence,
`k = c^2`, and since
`c` is an integer,
`k` must be a perfect square. Let's look at the
`x^2` coefficient again. We must have that
`2c - b^2 = 0`, so
`b^2 = 2c`. Hence,
`c` must itself be half of a perfect square, since
`b` is an integer.

Let's consider values of
`b`. Since
`b^2 = 2c`,
`b^2` is even, so
`b` is even. If
`b = 2`,
`c = 2`, and
`k = c^2 = 4`. If
`b = 4`,
`c = 8`, and
`k = 64`. If
`b = 6`,
`c = 18`, so
`k = 324`. If
`b = 8`,
`c = 32`, and
`k = 1024`. If
`b = 10`,
`c = 50`, so
`k = 2500`: but that doesn't work, because
`2500 > 2000`. So, our working values for
`k` are 4, 64, 324, and 1024.

This was a very drawn-out problem, so if you have any questions about how I worked through any of the steps, please feel free to ask!