For which integers $n$ does $$(t-n)^2+(t-(n-1))^2+(t-(n-2))^2+...+(t-1)^2=\sum_{i=1}^ni$$ have solutions?
I have managed to find an answer to this, but it is probably wrong (example later). Note that this is not a textbook question; rather one I've created.
Attempt
I will give a very brief answer to this.
The LHS can be written as $$\small(t^2-2nt+n^2)+((t^2-2nt+n^2)+2t-2n+1)+((t^2-2nt+n^2)+4t-4n+4)+...+((t^2-2nt+n^2)+2(n-1)t-2n(n-1)+(n-1)^2)$$ which can be simplified as $$n(t^2-2nt+n^2)+n(n-1)t-\frac16n(4n^2+9n-1)=n(t^2-(n-1)t+\frac16(2n^2-9n+1))$$ using the sum of squares. So our equation can be expressed as $$n(t^2-(n-1)t+\frac16(2n^2-9n+1))=\frac12n(n+1)$$ giving $$t^2-(n-1)t+\frac13(n^2-6n-1)=0$$ The equation has solutions if the discriminant is not less than $0$: $$b^2-4ac=(n-1)^2-\frac43n(n^2-6n-1)>0$$ when $n<7$ by Desmos.
(Counter)example
Let $n=7$. Then the equation becomes $$7t^2-56t+140=28\implies t=4$$ which is a solution. This contradicts my attempt and the result obtained by Desmos.
Can anyone spot the error in my answer?
This is correct.
This is incorrect. The last term has an error.
Since we have, for $1\le j\le n$, $$\begin{align}(t-j)^2&=(t-n+n-j)^2\\\\&=(t-n)^2+2(t-n)(n-j)+(n-j)^2\\\\&=(t-n)^2+2(n-j)t-n^2+j^2\end{align}$$ we have$$\begin{align}&\sum_{j=1}^{n}((t-n)^2+2(n-j)t-n^2+j^2)\\\\&=n(t-n)^2+2\cdot\frac{n(n-1)}{2}t-n^3+\frac{n(n+1)(2n+1)}{6}\\\\&=n(t^2-2nt+n^2)+n(n-1)t-\frac 16n(4n^2\color{red}{-3}n-1)\end{align}$$
We want to find integers $n$ such that $$n(t^2-2nt+n^2)+n(n-1)t-\frac 16n(4n^2-3n-1)=\frac{n(n+1)}{2}$$ has solutions.
This is equivalent to $$3t^2+(-3n-3)t+n^2-1=0$$ and so $$t=\frac{3n+3\pm\sqrt{-3n^2+18n+21}}{6}$$ So, we have to have $$-3n^2+18n+21\ge 0\implies -1\le n\le 7$$
Therefore, if $t\in\mathbb Z$, then the answer is $$(n,t)=(1,0),(1,2),(5,2),(5,4),(7,4)$$