It can be proven that $4^{2000} + 4^x + 4^{2023}$ is the square of an integer for exactly 3 integer values of $x$. What is the largest such value of $x$?
This is a problem from a high school mathematics competition. I factored out $4^{2000}$ from the expression to turn the expression into $(1 + 4^{x-2000} + 4^{23})4^{2000}$, which is only a perfect square if $1 + 4^{x-2000} + 4^{23}$ is a perfect square, but I am not sure what I could do from there to determine for what values of $x$ that expression is a perfect square
Assuming that there are exactly three integer values of $x$ for which $4^{2000} + 4^x + 4^{2023}$ is a square integer, the largest such value is $x = 2045$. You can work this out by noticing that the given expression can be written as $(2^{2000})^2+(2^x)^2 + (2^{2023})^2$ and so you can try combinations for which this is a perfect square $(a + b)^2$. Working out what $x$ can be then comes down to matching up the non-squared term, so $(2^x + 2^{2023})^2$ gives $x = 1976$, $(2^{2000} + 2^{2023})^2$ gives $x = 2012$, and finally $(2^x + 2^{2000})^2$ gives $x = 2045$.