Question 163472
Let's call the first integer, N.
the next 5 consecutive integers would be
{{{N+1}}}
{{{N+2}}}
{{{N+3}}}
{{{N+4}}}
{{{N+5}}}
and their squares would be,
{{{N^2=N^2}}}
{{{(N+1)^2=N^2+2N+1}}}
{{{(N+2)^2=N^2+4N+4}}}
{{{(N+3)^2=N^2+6N+9}}}
{{{(N+4)^2=N^2+8N+16}}}
{{{(N+5)^2=N^2+10N+25}}}
Now add them together,
{{{N+(N+1)^2+(N+2)^2+(N+3)^2+(N+4)^2+(N+5)^2=(N^2)+(N^2+2N+1)+(N^2+4N+4)+(N^2+6N+9)+(N^2+8N+16)+(N^2+10N+25)}}}
{{{N+(N+1)^2+(N+2)^2+(N+3)^2+(N+4)^2+(N+5)^2=6N^2+30N+55}}}
You know the sum of the squares equals 1111.
{{{6N^2+30N+55=1111}}}
{{{6N^2+30N-1056=0}}}
{{{N^2+5N-176=0}}}
{{{(N+16)(N-11)=0}}}
Two solutions.
.
.
.
{{{N-11=0}}}
{{{N=11}}}
The integers are then 11,12,13,14,15,16.
.
.
.
{{{N+16=0}}}
{{{N=-16}}}
The integers are then -16,-15,-14,-13,-12,-11.