Question

Another math problem. Can you solve it? I can't... For a good answer I'll make it 'The Best' I hope you can help me... Thanks

Answer 1

`\boxed{\sf \ \ \ 10^2+11^2+12^2=13^2+14^2 \ \ \ }`

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

`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\\`

as we are looking for positive integer the solution is a = 10

do not hesitate if you have any question