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#