# How to solve this math problem?Prove that the sum of squares of five consecutive integers can not be written as the sum of squares of three consecutive integers.

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### 2 Answers

We let the sum of 5 consecutive squares be represented as :

`(n-2)^2+(n-1)^2+n^2+(n+1)^2+(n+2)^2`

`n^2-4n+4+n^2-2n+1+n^2+n^2+2n+1+n^2+4n+4`

`5n^2+10`

We let the sum of 3 consecutive squares be represented as:

`(m-1)^2+m^2+(m+1)^2` or `3m^2+2` .

Note that we used two variables, as it is not necessary that the center number be the same.

If these sums are equal then:

`5n^2+10=3m^2+2`

`5n^2+8=3m^2` .

This is impossible since `5n^2+8` is never a multiple of 3 if `n` is an integer. Proof:

`n` is either 1 less than a multiple of 3, a multiple of 3, or 1 more than a multiple of 3. We examine each case:

(1) Let `n=3p-1` . Then

`5n^2+8=5(3p-1)^2+8=45p^2-30p+13=3(15p^2-10p+4)+1` which cannot be a multiple of 3.

(2) Let `n=3p`

`5n^2+8=5(3p)^2+8=45p^2+8=3(15p^2+2)+2` which cannot be a multiple of 3.

(3) Let `n=3p+1`

`5n^2+8=5(3p+1)^2+8=45p^2+30p+13=3(15p^2+10p+4)+1` which cannot be a multiple of 3.

Therefore `5n^2+8` is never a multiple of 3, and cannot equal `3m^2` .

So the sum of 5 consecutive squares cannot be written as a sum of 3 consecutive squares.

You should assume that the first integer is n, hence the next 4 consecutive integers is: `n+1,n+2,n+3,n+4`

You need to write the sum of squares of the 5 consecutive integers such that:

`n^2 + (n+1)^2 + (n+2)^2 + (n+3)^2 + (n+4)^2`

Expanding the binomials yields:

`n^2 + n^2 + 2n + 1 + n^2 + 4n + 4 +n^2 + 6n + 9 + n^2 + 8n + 16 = 5n^2 + 20n + 30`

You need to write the sum of squares of the 3 consecutive integers such that:

`n^2 + (n+1)^2 + (n+2)^2 = n^2 + n^2 + 2n + 1 + n^2 + 4n + 4 = 3n^2 + 6n + 5`

**Hence, evaluating the sum of squares of 5 consecutive integers and the sum of squares of 3 consecutive integers yields the difference between the equations representing these sums such that: `5n^2 + 20n + 30 != 3n^2 + 6n + 5` .**