Modular Arithmetic problem: Show that $\sum_{k=1}^{99}{(x+k)^2} = y^z$ is not solvable in integers x,y,z if z>1

Question

Expanding the LHS:

$\sum_{k=1}^{99}{(x+k)^2} = 99x^2 + 2(\sum_{k=1}^{99}{k})x + (\sum_{k=1}^{99}{k^2}) = 99x^2 + 9900x + 33\times50\times199$

Reducing it to (mod 99) gives:

$66 \equiv y^z \space\space$ (mod 99)

or Reducing it to (mod 33) (or any other factors) gives:

$0 \equiv y^z \space\space$ (mod 33)

Then I don't know how to show that $y^z$ has no solutions for $z>1$.

Could anyone offer a hint or a solution?

Answer 1

Note that

$$y^z = 99x^2 + 9900x + 33 \times 50 \times 199 = 3(3(11)x^2 + 3(1100)x + 11(50)(199))$$

Since the first $2$ terms in the brackets have a factor of $3$ but the third one does not, this indicates $y^z$ has exactly $1$ factor of $3$. Because $3$ is prime, we get that $3 \mid y^z \; \to \; 3 \mid y$ (e.g., using Euclid's lemma), i.e., $y$ has at least $1$ factor of $3$. Thus, $y^z$ has at least $z$ factors of $3$. However, since $y^z$ actually has only one factor of $3$, this means $z$ can't be more than $1$.

Answer 2

You have shown that $33|y^{z}$ and since $33 = 11 \cdot 3$ where both are prime,

if $z > 1$, then $33|y^{z} \implies 33|y \implies 99|y^{z}$ and checking $\pmod {99}$ on your above expression leaves

$y^{z} \equiv 0 \equiv 33 \cdot 50 \equiv 66\pmod{99}$ a contradiction.