91Ó°ÊÓ

The problem is related to the quadratic equation \(x^2 + 87559743x - 787727821\), and we need to prove that it has no integer solutions. We will do this by using the properties of even and odd numbers when they are added, multiplied or subtracted.

Let's break down the quadratic equation into its three components: 1. \(x^2\) 2. \(87559743x\) 3. \(-787727821\) Now we will analyze the even-odd cases for each component depending on if x is odd or even.

If \(x\) is odd, then multiplying it with an odd constant will result in another odd result because (odd * odd = odd). 1. \(x^2\): odd (since odd * odd = odd) 2. \(87559743x\): odd (since odd * odd = odd) 3. \(-787727821\): odd (since it is given as odd) Adding or subtracting odd numbers will result in even numbers. We are adding the first two components and subtracting the third, which means: (odd + odd) - odd = (even) - odd The result will be an odd number, whereas the right-hand side of the equation is an even number (0). Hence, there's no integer solution if \(x\) is odd.

If \(x\) is even, then squaring it will result in an even number because (even * even = even). However, multiplying it with an odd constant will give us an even result as well (even * odd = even). 1. \(x^2\): even (since even * even = even) 2. \(87559743x\): even (since even * odd = even) 3. \(-787727821\): odd (since it is given as odd) Adding or subtracting even numbers will yield even numbers, whereas adding or subtracting an odd number will yield an odd number. We are adding the first two components and subtracting the third component: (even + even) - odd = (even) - odd This time, the result will be an odd number. This contradicts the right-hand side of the equation being an even number (0). Therefore, there's no integer solution if \(x\) is even as well.

Since the given quadratic equation doesn't result in an integer solution for both odd and even values of \(x\), we can conclude that the equation \(x^2 + 87559743x - 787727821\) has no integer zeros.