Answer: i think the answer is
Let n, n+1, n+2, n+3 and n+4 be the 5 consecutive integers.
We are looking for n such that:
(n+4)²+(n+3)²=(n+2)²+(n+1)²+n²
Let n²+8n+16+n²+6n+9=n²+4n+4+n²+2n+1+n²
2n²+14n+25=3n²+6n+5
Let 3n²-2n²+6n-14n+5-25=0
n²-8n-20=0
Δ=8²+4*20=64+80=144
√Δ=12
The 2 solutions are n1=(8+12)/2=10 and n2=(8-12)/2=-2
As n is positive (because it is a natural number), there is only one possible solution: n=10
The 5 integers are therefore: 10, 11, 12, 13, 14.
We check: 13²+14²=169+196=365
10²+11²+12²=100+121+144=365
Explanation: