Is $k=3$ the only solution to $\sum_{r=1}^n r^k =\left( \sum_{r=1}^nr^1 \right)^2 $? $f(k) = \sum_{r=1}^{n} r^k$. Find an integer $x$ that solves the equation $f(x) = \bigl(f(1)\bigr)^2$.
Problem credit: http://cotpi.com/p/2/
I understand why $x = 3$ is a solution. $1^3 + 2^3 + \dots + n^3 = \left(\frac{n(n + 1)}{2}\right)^2 = (1 + 2 + \dots + n)^2$. But how can we prove that there is no other solution?
Will we have other solutions if $x$ is real? What if $x$ is complex?

I assume that you want to find values of $x$ for which $f(x) = (f(1))^2$ as functions in $n$. In this case, we can simply substitute in $n=1$, $n=2$, and $n=3$. Namely, we need to satisfy $$1^x=1 \qquad \text{and} \qquad 1^x + 2^x = 9 \qquad \text{and} \qquad 1^x + 2^x + 3^x = 36$$ Subtracting successive equations yields $$2^x = 8 \qquad \text{and} \qquad 3^x = 27$$ Obviously, $x = 3$ is the only real value which works. You might wonder whether there are any complex values. Suppose $x=a+bi$. Then $$8=2^x = 2^a2^{bi} = 2^ae^{ib\ln{2}}$$ Matching magnitudes, we need $2^a=3$, so that $a = 3$. Meanwhile, notice that $b$ must take the form $\frac{2\pi}{\ln{2}}m$ for some integer $m$. If we continue this analysis for the equation $3^x=27$, we also need that $b = \frac{2\pi}{\ln{3}}n$ for some integer $n$. Setting these expressions for $b$ equal and solving yields $$m \ln{3} = n\ln{2},$$ or equivalently, $$3^m = 2^n,$$ which is only true when $m=n=0$. Therefore, $b=0$, and so $x=3$ is the only value for which $f(x) = (f(1))^2$ for every $n$.
