Show that if x,y,z are positive integers, then (xy+1)(yz+1)(zx+1) is a perfect square if and only if xy+1,yz+1,zx+1 are all perfect squares.

Source: the probem posed and resolved by Euler.

Consider small values with x=1,2,3...

1 3 8
1 8 15
1 15 24
1 24 35...

2 4 12
2 12 24
2 24 40...

3 5 16
3 16 33
3 33 56...

(there are also solutions

3 1 8
3 8 21
3 21 40..., but as this is just a partial proof, these can be left for another occasion).

With a little work, it is readily seen that the recurrence formula for

x=1 is y = (n+1)^2-1, z=((n+1)+1)^2-1
x=2 is y = 1/2(2n +1)^2-1/2, z=1/2((2n+1)+1)^2-1/2
x=3 is y = 1/3(3n+1)^2-1/3,  z=1/3((3n+1)+1)^2-1/3, etc.

From these hints it is easily to compute, for x=k:

xy+1=k*(1/k((k*n+1)^2-1))+1 = (kn+1)^2    
xz+1=k*(1/k((k*(n+1)+1)^2-1))+1  = (k(n+1)+1)^2    
yz+1= (1/k((k*n+1)^2-1))*(1/k((k*(n+1)+1)^2-1))+1 = (kn^2 + kn + 2n + 1)^2, all squares, for any {n,k} as was to be shown.

Then for (xy+1)(yz+1)(zx+1) we have the square of S, where  S=(kn+1) (k(n+1)+1)(kn^2+kn+2n+1)

k=1: 30,132,380,870,1722,3080, 5112,8010,11990,...               
k=2: 105,595,1953,4851,10153,18915,32385,52003,79401,...
k=3: 252,1610,5590,14352,30704,58102,100650,163100,250852,… etc.

Edited on March 28, 2018, 6:57 am

Posted by broll on 2018-03-28 06:50:15