Question #eb541

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= (3^2 cdot 2)^2` the possible factors for `(3^2 cdot 2)^2` are

`f={1,2,3,4,6,9,18,36,81,108,162,324}`

Now solving for `n,m` the linear systems

`{(m+n=f_i),(m-n=324/f_i):}`

for `i=1,2,cdots,12`

we get the `n,m` numbers

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`.

`n+18 = b^2` and `n-18 - a^2`

Then `b^2-a^2 = 36` with `a`, `b` integers.

Therefore,

`(b+a)(b-a) = 36` with `a`, `b`, `b+a` and `b-a` all integers.

the integer factorizations of `36` are

`{:(b+a," ",b-a),(36," ",1),(18," ",2),(12," ",3),(9," ",4),(6," ",6) :}`

Because `b` must be an integer , the sum of the two numbers, which is `2b`, 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=20/2 = 10`, so `b^2 = 100` and `n=b^2-18 = 82`

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

`b= 12/2 = 6`, so `b^2 = 36` and `n=b^2-18 = 18`