Solving diophantine equations: Finding integer solutions: $(x+1)^2+(x+2)^2+...+(x+2001)^2=y^2$

Question

I would like to find all integer solutions to the Diophantine equation. I wonder if there is a way to solve it using Fermat.

$(x+1)^2+(x+2)^2+\ldots+(x+2001)^2=y^2$

Answer 1

I think your problem does not have a solution.

To see this, notice that $x^2 \equiv 1 \ \text{mod} \ 3$ whenever $x$ is not divisible by $3$. Further, $2001$ is divisible by $3$, thus, on the right hand side you have $$\sum_{i=1}^{2001} (x+i)^2 \equiv \frac{2001}{3} \sum_{i=1}^{3} (x+i)^2 \equiv \frac{2001}{3} \sum_{i=1}^{3} i^2 \equiv 1334 \equiv 2 \quad \text{mod} \ 3$$ while the left hand side gives remainder $0$ or $1$ mod $3$.

As per your question, Fermats (presumably little) theorem says that a prime $p$ and $a$ not a multiple of $p$, $a^{p-1} \equiv 1 \ \text{mod} \ 3$, which would give you the remainders I talked about above. In my opinion this is a bit overkill for the question.

If you fancy, you can also generalise this argument to many further cases. You might wish to establish which other numbers instead of $2001$ would this argument still go trough. You can also estabish similar equations for powers other than 2, i.e. the solvability of an equation of the form $$\sum_{i=1}^{N}(x+i)^p = y^p$$ this way maybe making a bit more use of Fermat's little theorem.