How to solve sum of quadratics

Question

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?

"which can be simplified as" then coefficient of $t$ should be $n(n+1)$ — Will Jagy, Jan 24 '18 at 20:38
@WillJagy But we have $$2t+4t+...+2(n-1)t=2t(1+2+...+(n-1))=2t\cdot\frac{(n-1)((n-1)+1)}2=n(n-1)t$$ — ə̷̶̸͇̘̜́̍͗̂̄︣͟, Jan 25 '18 at 07:31
the expression shaded yellow begins with $(t-n)^2$ and goes to $(t-1)^2$ — Will Jagy, Jan 25 '18 at 17:52
The right hand side of the equation highlighted in yellow should be $\displaystyle\sum_{i=1}^n(t-i)^2$ — robjohn, Jan 26 '18 at 16:25

mathlove · Accepted Answer · 2018-01-26T16:20:23.997

The LHS can be written as $$\small(t^2-2nt+n^2)+((t^2-2nt+n^2)+2t-2n+1)+((t^2-2nt+n^2)+4t-4n+4)+...+((t^2-2nt+n^2)+2(n-1)t-2n(n-1)+(n-1)^2)$$

This is correct.

which can be simplified as $$n(t^2-2nt+n^2)+n(n-1)t-\frac16n(4n^2+9n-1)$$

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 $$(n,t)=(1,0),(1,2),(5,2),(5,4),(7,4)$$

How to solve sum of quadratics

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