# How do you find three consecutive odd integers such that the product of the two smaller exceeds the largest by 52?

Jan 12, 2017

$7 , 9 , 11$

#### Explanation:

Note the question speaks of "smaller" and "largest" numbers rather than "lesser" and "greatest", so it is easier to split positive and negative cases...

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Case - positive integer solutions

Considering positive integers first, let the smallest be denoted by $n$, so the other two are $n + 2$ and $n + 4$.

We are given:

$n \left(n + 2\right) = \left(n + 4\right) + 52$

which expands to:

${n}^{2} + 2 n = n + 56$

Subtract $n + 56$ from both sides to get:

$0 = {n}^{2} + n - 56 = \left(n + 8\right) \left(n - 7\right)$

Which gives us $n = 7$ or $n = - 8$.

Since we specified positive integers, the three numbers are $7 , 9 , 11$

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Case - negative integer solutions

Now consider negative integers.

If the smallest negative integer is $n$, then the other two are $n - 2$ and $n - 4$.

We are given:

$n \left(n - 2\right) = \left(n - 4\right) + 52$

which expands to:

${n}^{2} - 2 n = n + 48$

Subtract $n + 48$ from both sides to get:

$0 = {n}^{2} - 3 n - 48$

Multiply by $4$ so we can complete the square with integers:

$0 = 4 \left({n}^{2} - 3 n - 48\right)$

$\textcolor{w h i t e}{0} = 4 {n}^{2} - 12 n - 192$

$\textcolor{w h i t e}{0} = {\left(2 n\right)}^{2} - 2 \left(2 n\right) \left(3\right) + 9 - 201$

$\textcolor{w h i t e}{0} = {\left(2 n + 3\right)}^{2} - 201 \text{ }$ (I could stop here, but...)

$\textcolor{w h i t e}{0} = {\left(2 n + 3\right)}^{2} - {\left(\sqrt{201}\right)}^{2}$

$\textcolor{w h i t e}{0} = \left(\left(2 n + 3\right) - \sqrt{201}\right) \left(\left(2 n + 3\right) + \sqrt{201}\right)$

$\textcolor{w h i t e}{0} = \left(2 n + 3 - \sqrt{201}\right) \left(2 n + 3 + \sqrt{201}\right)$

$\textcolor{w h i t e}{0} = 4 \left(n + \frac{3}{2} - \frac{\sqrt{201}}{2}\right) \left(n + \frac{3}{2} + \frac{\sqrt{201}}{2}\right)$

So $n = - \frac{3}{2} \pm \frac{\sqrt{201}}{2} \text{ }$ (not an integer)

There are no (negative) integer solutions since $201$ is not a perfect square.