Each of x, y, and z is a
positive integer.
Derive a parametric formula that provides all possible solutions to this equation:
 (x^{2}+x+1)(y^{2}+y+1) = (z^{2}+z+1)
If not for the positive integer restriction, (x,y,z) = (0,t,t) or (t,0,t) would work, but x,y,z must be non zero, so scratch that one.
After some algebra, one parametric solution is:
(x,y,z) = (t1, t, t^2)
(x,y,z) = (t, t1, t^2) (equivalent since x and y are interchangeable)
Proof:
Consider the substitution of y = (x+1)
(x^2+x+1)*((x+1)^2+(x+1)+1) =
(x^2+x+1)*(x^2 + 2x + 1+x+2) =
(x^2 + x + 1)*(x^2 + 3x + 3) =
x^4 + 4x^3 + 7x^2 + 6x + 3 =
(x^4 + 4x^3 + 6x^2 + 4x + 1) + (x^2 + 2x + 1) + 1
(x+1)^4 + (x+1)^2 + 1
letting z = (x+1)^2 yields z^2+z+1
But wait, there's more.
Using a program, (setting WLOG x <= y), I found several other solutions but I have not been able to put them into a parametric form. Note that I did not find any other solutions other than the above parametric formula for any x larger than 11, but that might not hold if larger y and z values were allowed.
x y z
1 2 4
1 9 16
1 35 61
1 132 229
1 494 856
2 3 9
2 11 30
2 25 67
2 56 149
2 183 485
2 406 1075
3 4 16
3 49 178
3 87 315
3 305 1101
4 5 25
4 26 121
4 81 373
4 390 1789
5 6 36
5 84 470
5 121 676
6 7 49
6 169 1111
6 230 1511
7 8 64
7 74 562
7 225 1702
8 9 81
9 10 100
9 55 529
9 68 653
10 11 121
10 146 1543
11 12 144
11 129 1493
12 13 169
13 14 196
14 15 225
t1 t t^2 afterwards
Putting in the y values when x=1: 2,9,35,132,494 gives oeis A001571
a(n) = 4*a(n1)  a(n2) + 1, with a(0) = 0, a(1) = 2
While putting in the z values when x=1: 4,16,61,229,856 gives oeis A133161
Indices of the triangular numbers which are also centered triangular number.
This is not a complete solution. For one thing, allowing larger values for y and z will find more solutions. For another, I don't have a parametric version for these other solutions.

Posted by Larry
on 20220527 08:51:20 