Question

Three consecutive odd integers are such that the square of the third integer is 345 less than the sum of the squares of the first two. How do you find the integers?

Answer 1

There are two solutions:

`21, 23, 25`

or

`-17, -15, -13`

Explanation:

If the least integer is `n`, then the others are `n+2` and `n+4`

Interpreting the question, we have:

`(n+4)^2 = n^2+(n+2)^2-345`

which expands to:

`n^2+8n+16 = n^2 + n^2+4n+4 - 345`

`color(white)(n^2+8n+16) = 2n^2+4n-341`

Subtracting `n^2+8n+16` from both ends, we find:

`0 = n^2-4n-357`

`color(white)(0) = n^2-4n+4-361`

`color(white)(0) = (n-2)^2-19^2`

`color(white)(0) = ((n-2)-19)((n-2)+19)`

`color(white)(0) = (n-21)(n+17)`

So:

`n = 21" "` or `" "n = -17`

and the three integers are:

`21, 23, 25`

or

`-17, -15, -13`

`color(white)()`
Footnote

Note that I said least integer for `n` and not smallest.

When dealing with negative integers these terms differ.

For example, the least integer out of `-17, -15, -13` is `-17`, but the smallest is `-13`.