College

Solve for [tex]$x$[/tex] and [tex]$y$[/tex] if they are integers.

[tex]$4x^4 + 5y^4 + 45x^2y^2 = 0$[/tex]

Answer :

To solve the given equation [tex]\(4x^4 + 5y^4 + 45x^2y^2 = 0\)[/tex] for integers [tex]\(x\)[/tex] and [tex]\(y\)[/tex], we need to find combinations of integers that satisfy this equation.

Let's analyze the structure of this equation. The equation can be rewritten as:

[tex]\[
4x^4 + 45x^2y^2 + 5y^4 = 0
\][/tex]

Notice that each term involves squares and fourth powers, suggesting that we are dealing with a form that represents a sum of squares. For the sum of these terms to be zero, each of these terms essentially needs to cancel one another out.

### Step 1: Consider the possibility of [tex]\(x\)[/tex] or [tex]\(y\)[/tex] being zero.

1. If [tex]\(x = 0\)[/tex]:

- The equation becomes [tex]\(5y^4 = 0\)[/tex].
- Solving this, we find [tex]\(y = 0\)[/tex].

2. If [tex]\(y = 0\)[/tex]:

- The equation becomes [tex]\(4x^4 = 0\)[/tex].
- Solving this, we find [tex]\(x = 0\)[/tex].

From these results, one solution is [tex]\((x, y) = (0, 0)\)[/tex].

### Step 2: Consider non-trivial cases where neither [tex]\(x\)[/tex] nor [tex]\(y\)[/tex] is zero.

Since both variables are non-zero, let's factor the equation assuming non-zero values:

The equation remains intricate, and without complex or non-integer numbers, finding solutions is challenging. The symmetry and powers involved suggest that any proportionality in terms of non-zero [tex]\(x\)[/tex] and [tex]\(y\)[/tex] cannot satisfy the equation with real integer solutions since they introduce complex numbers when broken down.

### Conclusion:

After analyzing the equation and considering possible integer values, the only integer solution to the equation remains [tex]\((x, y) = (0, 0)\)[/tex]. For all other cases, ensuring that the equation evaluates to zero would mean involving non-real numbers, which are not integers. Therefore, the only feasible integer solution is:

[tex]\[
(x, y) = (0, 0)
\][/tex]