Problem (Algebra):
There are two increasing sequences of five consecutive integers, each of which have the property that the sum of the squares of the first three integers in the sequence equals the sum of the squares of the last two. Determine these two sequences.

Solution:
Let $n$ be the smallest integer in one of these sequences.

So we want to solve the equation $n^{2}+(n+1)^{2}+(n+2)^{2}=(n+3)^{2}+(n+4)^{2}$ (translating the given problem into an equation).

Thus $n^{2}+n^{2}+2 n+1+n^{2}+4 n+4=n^{2}+6 n+9+n^{2}+8 n+16$



$$
\begin{array}{r}
n^{2}-8 n-20=0 \\
(n-10)(n+2)=0
\end{array}
$$

So $n=10$ or $n=-2$.

Therefore, the sequences are 10, 11, 12, 13, 14 and $-2,-1,0,1,2$.

Verification 

$(-2)^{2}+(-1)^{2}+0^{2}=1^{2}+2^{2}=5$ and $10^{2}+11^{2}+12^{2}=13^{2}+14^{2}=365$

Final answer: 10,11,12,13,14,-2,-1,0,1,2