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

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

Let the trinomial factors be and . Then, their product is . We'll solve this problem by analyzing each of these coefficients.

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

So, we're back to and . We have that the coefficient is , and since that term is zero in this polynomial, we have . This means that .

Our polynomials are now and . Let's consider the coefficient. It shows that we must have , so either or .

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

Our polynomials are thus and . Hence, , and since is an integer, must be a perfect square. Let's look at the coefficient again. We must have that , so . Hence, must itself be half of a perfect square, since is an integer.

Let's consider values of . Since , is even, so is even. If , , and . If , , and . If , , so . If , , and . If , , so : but that doesn't work, because . So, our working values for 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!

step-by-step explanation:

idk the answer i want to say 10 but then it would be equal so try 9 but then yeah so yeah we love edgen