Too Much Fractions

To make things a bit nicer, multiply both sides of the equation by $61^x$ . Hence, the equation becomes,

$55^x + 56^x +$ $...$ $+ 60^x =$ $61^x + 62^x +$ $... + 65^x$

By doing a little bit of Trial and Error, we get that $\boxed{x = 2}$ is a solution. (Well, that is if you're not lazy.)

But how do we get $x = 2$ ?

Consider the identity,

$a^2 + (a+1)^2 +$ $... + (a+k)^2 =$ $(a+k+1)^2 +$ $... + (a+2k)^2$

for $a = k(2k+1), k \in$ $\mathbb{N}$

The equation can be rewritten as,

$55^x + (55+1)^x +$ $...$ $+ (55+5)^x =$ $(55+6)^x + (55+7)^x +$ $... + (55+$ $10$ $)^x$

Then, $k = 5$ and indeed, $a = 5\times(2\times5 + 1) = 55$

Since the equation follows the identity, we know now (at least not by Trial and Error) that $x = 2$ is a solution.

To prove that only $x = 2$ is the solution, notice in the original equation that the left-hand side is a decreasing function in $x$ and the right-hand side is an increasing function in $x$ , thus, it has at most one solution, which is $\boxed{x=2}$ .

Therefore, the value of $(x^2+1)^4 + (x+23)^2 + (x+1)^6 + 36 = (4+1)^4 + (2+23)^2 + (2+1)^6 + 36 = 625 + 625 + 729 + 36 = \boxed{2015}$

This problem is modified from Zuming Feng and Titu Andreescu's book, 101 Problems in Algebra From the Training of the USA IMO Team