Question

If the sum of the squares of three consecutive integers is 194, what are the numbers?

Answer 1

`7,8,9` or `-9,-8,-7`

Explanation:

`a^2 + b^2 + c^2 = 194`

There are two possible scenarios, either the three numbers go even-odd-even or they go odd-even-odd. Any even number can be written as `2k, kinZZ` and any odd number can be written as `2j+1, jinZ`. For the first scenario, we have:

`(2k)^2 + (2k+1)^2 + (2(k+1))^2 = 194`

`(2k)^2 + (2k+1)^2 + (2k+2)^2 = 194`

`4k^2 + 4k^2 + 4k + 1 + 4k^2 + 8k + 4 = 194`

`12k^2+12k+5 = 194`

We now have a quadratic we can solve for k.

`12k^2 + 12k - 189 = 0`

`k = (-12+-sqrt(144-4(12)(-189)))/24 = 7/2 or -9/2`

These are not integer values so this is not the case.

Trying the second case, we have:

`(2k+1)^2 + (2(k+1))^2 + (2(k+1)+1)^2 = 194`

`(2k+1)^2 + (2k+2)^2 + (2k+3)^2 = 194`

`4k^2+4k+1 + 4k^2+8k+4 + 4k^2+12k+9 = 194`

`12k^2+24k+14=194`

`12k^2+24k-180=0`

`12(k^2+2k-15) = 0`

`12 != 0 implies k^2+2k-15=0`

Factorising our quadratic gives;

`(k+5)(k-3) = 0`

so k is 3 or -5. This means our numbers are:

`7,8,9` or `-9,-8,-7`

Answer 2

Three consecutive integers are `{-9,-8,-7}` or `{7,8,9}`.

Explanation:

Let the three numbers be `x-1`, `x` and `x+1`. As sum of their squares is `194`, we have

`(x-1)^2+x^2+(x+1)^2=194` or

`x^2-2x+1+x^2+x^2+2x+1=194` or

`3x^2+2=194` or `3x^2-192=0`

Or `x^2-64=0` i.e. `(x+8)(x-8)=0`

Hence `x=-8` or `x=8` and as this is middle number

Three consecutive integers are `{-9,-8,-7}` or `{7,8,9}`.