Question

Prove that the sum of the squares of 5 consecutive numbers is divisible by 5.

Begin by writing the middle number as n, so that the other numbers are n-2, n-1, n +1, n+2

How would I prove this? I tried to but ended with n^10 x 10 and didn't know if that's even divisble by 5 or how it to say it divides by 5

Any help greatly appreciated! 10pts will be given out

## Answers (3)

Sum = (n-2)² + (n-1)² + n² + (n+1)² + (n+2)²

Sum = n² - 4n +4 + n² - 2n + 1 + n² + n² + 2n + 1 + n² + 4n + 4

Sum = (n² + n² + n² + n² + n²) + (-4n - 2n + 2n + 4n) + (4 + 1 + 1 + 4)

Sum = 5n² + 10

Sum = 5(n² + 2)

Since the factors of the sum includes 5, the sum is obviously divisible by 5.

Sum/5 = n² + 2

...

x²+(x+1)²+(x+2)²+(x+3)²+(x+4)²= 5x²+20x+30= 5(x²+4x+6)

start with n

(n)+(n+1)(n+2)(n+3)(n+4)(n+5)

5n+15

5(n+3)

which is divisible by 5