Question #eb541

Feb 9, 2017

See below.

Explanation:

If $m - 18$ and $m + 18$ are both perfect squares their product

${m}^{2} - {18}^{2} = {n}^{2}$ also is a perfect square.

Making now ${m}^{2} - {n}^{2} = {18}^{2} = {\left({3}^{2} \cdot 2\right)}^{2}$ the possible factors for ${\left({3}^{2} \cdot 2\right)}^{2}$ are

$f = \left\{1 , 2 , 3 , 4 , 6 , 9 , 18 , 36 , 81 , 108 , 162 , 324\right\}$

Now solving for $n , m$ the linear systems

$\left\{\begin{matrix}m + n = {f}_{i} \\ m - n = \frac{324}{f} _ i\end{matrix}\right.$

for $i = 1 , 2 , \cdots , 12$

we get the $n , m$ numbers

Feb 9, 2017

The only such integers are $18$ and $82$

Explanation:

Consider the fact that the perfect squares are generated by successive addition of the odd integers.

Image By Aldoaldoz - Own work, CC BY-SA 3.0, https://commons.wikimedia.org/w/index.php?curid=9757461

We have been asked to find all pairs of perfect squares whose difference is $36$

A bit of algebra will show that if the sum of two consecutive odds is $36$, then the two numbers are $17$ and $19$

This makes the squares ${8}^{2} = 64$ and ${10}^{2} = 100$.
The middle number is $82$ and $82 - 18 = 64$ and $82 + 18 = 100$

There are no four consecutive odd integers that sum to 36.

The first six consecutive odds sum to $36$.
This gives us the other solution of ${0}^{2} = 0$ and ${6}^{2} = 36$.
The middle number is $18$ and $18 - 18 = 0$ and $18 + 18 = 36$

It is not possible to find eight or more consecutive odds that sum to $36$.

Feb 9, 2017

And another solutions.

Explanation:

$n + 18 = {b}^{2}$ and $n - 18 - {a}^{2}$

Then ${b}^{2} - {a}^{2} = 36$ with $a$, $b$ integers.

Therefore,

$\left(b + a\right) \left(b - a\right) = 36$ with $a$, $b$, $b + a$ and $b - a$ all integers.

the integer factorizations of $36$ are

$\left.\begin{matrix}b + a & \text{ " & b-a \\ 36 & " " & 1 \\ 18 & " " & 2 \\ 12 & " " & 3 \\ 9 & " " & 4 \\ 6 & " } & 6\end{matrix}\right.$

Because $b$ must be an integer , the sum of the two numbers, which is $2 b$, must be even. (And $b$ is half of the sum.)

By exhaustion, the only two possibilities are

first
$b + a = 18$ and $b - a = 2$, so that

$b = \frac{20}{2} = 10$, so ${b}^{2} = 100$ and $n = {b}^{2} - 18 = 82$

second
$b + a = 6$ and $b - a = 6$, so that

$b = \frac{12}{2} = 6$, so ${b}^{2} = 36$ and $n = {b}^{2} - 18 = 18$