How do you find two consecutive positive odd integers whose product is 143?

Notice that consecutive odd integers will differ by `2`, so we could call the even number between them `n`, making the two integers `n-1` and `n+1`.

So:

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

Add `1` to both ends and transpose to get:

`n^2 = 144 = 12^2`

So `n = +-12`

Since we are told that the integers are positive, use `n=12` to find the pair of integers: `11, 13`.

Alternatively, just find factors of `143`, trying prime numbers in turn: `2` (no), `3` (no), `5` (no), `7` (no), `11` (yes). Then `143/11 = 13` - we have found our pair of numbers.

Proceed as below.
Let `(2n+1)` be the first odd integer, where `n` is any positive integer. Second number being consecutive is `(2n+3)`
Rest of the steps are same as followed by others.

It is by convention, as `2n+1` is always odd.

We need to be able to guarantee that we are dealing with odd numbers.

Let `n` be any number where `n>0`

Suppose n is even. Then `2n` is also even.
Suppose n is odd. Then `2n` is also even.

But in both cases `2n+1` will be odd.

Let the first odd number be `2n+1`
Then the second odd number will be `2n+3`

We are given that the product is 143 so write the equation as:

`(2n+1)(2n+3)=143`

This is saying exactly the same thing as the others but in a slightly different way! So we now have:

`4n^2+8n-140=0`

Divide by 4

`n^2+2n-35" "=" "(n+7)(n-5)" "=" "0`

So for my conditions `n=-7" or "+5`

`n=-7" does not work so we use " n=5`

Thus the first number is: `(2n+1)" "->" "2(5)+1=11`

So the second number is `11+2=13`