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Three squares imply the fourth (Posted on 2018-03-18) Difficulty: 4 of 5
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.

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How it goes | Comment 1 of 2
Diophantine tripla: a tripla of positive integers xyz whose products xy, xz, yz are less 1 of being a square and the product (xy+1)(yz+1)(zx+1) is a square.

Diophantine quadrupla: the same but with four numbers.


Euler? demonstrates that for each D-tripla there is also another positive integer w so that xw, yw, zw are less 1 of being a square. So that xyzw is a D-quadrupla.

I. e. (1,3,8) and (1,3,8,120)

Knowing the D-tripla it is possible to calculate the forth member to have a D-quadrupla (or to extend from xyz to xyzw) with the formula: 

(w+,w-)     w=x²+y²+z²+2xyz±(xy+1)(xz+1)(yz+1)
(note that w+ e w- are always integers)

in which w+,w- are the roots (two roots) of the quadratic equation:

x²+y²+z²+w²-2(xy+xz+xw+yz+yw+zw)-4xyzw-4=0

When xyz is a D-tripla the inferior root  (w-) is 0, and the superior root (w+) is w: the four value of the D-quadrupla xyzw.

But what if xyz is not a D-tripla (not being squares xy+1... ) but nonetheless the product (xy+1)(xz+1)(yz+1) is an square? Is it possible? Is it also extensible to a quadrupla xyzw? Euler shows that none of this was possible. The product of non-squares xy+1,.... is not an square.

The demostration goes like this: if some of the elements (xy+1), (xz+1), (yz+1) are not squares (and then xyz is not a D-tripla) but their product is a square, it is posible to find a forth element w- so that the product (xy+1)(xw-+1)(yw-+1) is a square and xyw- is not D-tripla. Note that the value of w- is not 0, as it would be if xyz was to be a D-tripla. 

w- is, instead, a positive integer and w-<z. 

Follows that we have got the product of three non-squares (xy+1)(xw+1)(yw+1) resulting in an square but with inferior value, because w-<z.

But then we are again in the initial conditions and we can repeat the process to get again an inferior value, and this infinitely many times: infinite descent.

There are three steps to demonstrate this: 

1) (xy+1)(xw-+1)(yw-+1) is a square and some of the factors are not square
2) w- is a positive integer
3)w- <z

Then xyw- is inferior than xyz but with the same properties. The infinite descent has begun.

Once understood this the demonstration is quite straightforward and it is possible to find in 

http://www.jstor.org/stable/2691347#page_scan_tab_contents

Edited on March 21, 2018, 6:57 pm
  Posted by armando on 2018-03-21 17:31:18

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