Answer:
[tex]\boxed{\sf \ \ \ 10^2+11^2+12^2=13^2+14^2 \ \ \ }[/tex]
Step-by-step explanation:
Hello,
let's note a a positive integer
5 consecutive integers are
a
a+1
a+2
a+3
a+4
so we need to find a so that
[tex]a^2+(a+1)^2+(a+2)^2=(a+3)^2+(a+4)^2\\<=>\\a^2+a^2+2a+1+a^2+4a+4=a^2+6a+9+a^2+8a+16\\<=>\\3a^2+6a+5=2a^2+14a+25\\<=>\\a^2-8a-20=0\\<=>\\(a+2)(a-10)=0\\<=>\\a = -2 \ or \ a = 10\\[/tex]
as we are looking for positive integer the solution is a = 10
do not hesitate if you have any question
