The sum of squares of five consecutive natural numbers is a multiple of 365, Give an example and prove it
10 ^2+11^2+12^2+13^2+14^2=365*2
Let the middle number be x, then
(x-2)^2+(x-1)^2+x^2+(x+1)^2+(x+2)^2
=5x^2+10
=5(x^2+2)
Let its value be 365 K and K be an integer
Then 5 (x ^ 2 + 2) = 365k = 5 * 73k
As long as x ^ 2 + 2 = 73k
Obviously, when x = 12, k = 2 satisfies
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