*sort by* **best** latest

### Best Answer I'M BANNED Y'ALL!!!!!!!!!! says

Using Rahul's answer, you can come up with infinitely more just by multiplying each number by the same sqaure number e.g. 4, 9, 16, 25, ... etc. The basic solution (2, 3, 6) generates the solutions

(8, 12, 24), (18, 27, 54), (32, 48, 96), (50, 75, 150) etc.

Allowing for two numbers to be identical, I also discovered a nice pattern of solutions of the form (1, 1, n) for n = 4, 144, 4900, 166464, 5654884, 192099600... The values of n have to be square numbers, and the sequence of square roots is 2, 12, 70, 408, 2378, 13860,... This square root sequence satisfies a linear recurrence: If A(k) is the kth term in the square root sequence, then

A(k) = 6*A(K-1) - A(k-2)

To generate the next solution, I compute 6*13860 - 2378 = 80782 and 80782^2 = 6525731524. Thus the next solution is (x, y, z) = (1, 1, 6525731524). (You can verify that it satisfies the conditions xy+xz+yz = square and xyz = square.)

interesting theorum here....

- See all 2 comments

### Jessee R says

One triplet is 2,3,6... by hit and trial... Unable to find a pattern or algebraic method for this one...

Perhaps... will need some more time

### Aficionada says

Are X, Y, and Z all different? That is, must they be?

And, are there definitely any solutions other than the one Rahul gave?

I have found a couple that come close, but do not fulfill the conditions. The ones I found both used two factors that were identical.

Brute force may be the only approach, but I assume all of us are starting with known perfect squares and working back from them? Also, some can be discarded without investigation, such as the square of a prime number, since it can't have 3 factors.

- See all 2 comments

Gave me an idea of another pattern: (1,4,9), (1,9,16), (1,16,25), (1,25,36), ... (1, n^2, (n+1)^2)... All these satisfy the conditions too.

- See all 2 comments

1 answer hidden due to negative feedback. Show