Let $$m$$ be the smallest positive integer such that $$m^{2}+(m+1)^{2}+\cdot\cdot\cdot+(m+10)^{2}$$ is the square of a positive integer $$n$$. Find $$m+n$$.

The sum of squares of 11 consecutive integers beginning with $$m$$ is

$$S=\sum_{k=0}^{10}(m+k)^2=\sum_{k=0}^{10}\bigl(m^2+2mk+k^2\bigr).$$

Using $$\sum_{k=0}^{10}k=55$$ and $$\sum_{k=0}^{10}k^{2}=385$$, we get

$$S=11m^{2}+2m\cdot55+385=11m^{2}+110m+385.$$

Factorising the constant 11,

$$S=11\bigl(m^{2}+10m+35\bigr) =11\bigl((m+5)^{2}+10\bigr).$$

We need $$S=n^{2}$$ for some positive integer $$n$$, so

$$n^{2}=11\bigl((m+5)^{2}+10\bigr).$$

Because 11 is prime, the bracketed term itself must be a multiple of 11. Put $$p=m+5$$ and demand

$$p^{2}+10=11s^{2} \qquad (s\in\mathbb Z^{+}). \; -(1)$$

First obtain the congruence condition modulo 11:

$$p^{2}+10\equiv0\pmod{11}\;\Longrightarrow\;p^{2}\equiv1\pmod{11}.$$

Thus $$p\equiv\pm1\pmod{11}\; \Longrightarrow\; p=11k\pm1\;(k\ge0).$$

Case 1: $$p=11k+1$$

Then from (1)

$$p^{2}+10=(11k+1)^{2}+10 =121k^{2}+22k+11 =11\bigl(11k^{2}+2k+1\bigr).$$

Therefore $$s^{2}=11k^{2}+2k+1. \; -(2)$$

Case 2: $$p=11k-1$$

Similarly,

$$p^{2}+10=(11k-1)^{2}+10 =121k^{2}-22k+11 =11\bigl(11k^{2}-2k+1\bigr)$$

so

$$s^{2}=11k^{2}-2k+1. \; -(3)$$

We need the right-hand sides of (2) or (3) to be perfect squares. Test small non-negative integers $$k$$ until a square appears (we also keep $$m=p-5>0$$):

k=0 gives $$m=-4$$ or $$m=-6$$ (negative, reject).

k=1:   Case 1 $$s^{2}=11+2+1=14$$ (not a square); Case 2 $$s^{2}=11-2+1=10$$ (not a square).

k=2:   Case 1 $$s^{2}=11\cdot4+2\cdot2+1=44+4+1=49=7^{2}.$$   Case 2 $$s^{2}=11\cdot4-2\cdot2+1=44-4+1=41$$ (not a square).

Thus the first success is in Case 1 with $$k=2$$, giving

$$p=11k+1=23,\quad m=p-5=18,\quad s=7.$$

Finally, $$n=11s=11\cdot7=77.$$

Hence $$m+n=18+77=95.$$

Answer: 95