1.Introduction

A problem of three squares whose differences are squares, i.e. is there a solution of simultaneous equation
{x^2-y^2=u^2, x^2-z^2=v^2, y^2-z^2=w^2}. 
According to Dickson's book[1], W.Lenhart gave two parametric solutions below.
x=(a^2+b^2)/(a^2-b^2), y=(c^2+d^2)/(c^2-d^2), z=1,
where {a=8n, b=n^2+9, c=8n(n^2-9), d=n^4+2n^2+81} or {a=n^2-3, b=2n, c=(n^2+1)^2+8, d=2n(n^2-3)}.

We show simultaneous equation {x^2-y^2=u^2, x^2-z^2=v^2, y^2-z^2=w^2} has infinitely many parametric solutions.
     
2.Theorem
     
A simultaneous equation {x^2-y^2=u^2, x^2-z^2=v^2, y^2-z^2=w^2} has infinitely many parametric solutions.
 
Proof.

x^2-y^2=u^2.........................................................................(1)

x^2-z^2=v^2.........................................................................(2)

y^2-z^2=w^2.........................................................................(3)

From equation (1), substitute x=m^2+n^2, y=m^2-n^2, and u=2mn to equation (2) and (3).

From equation (2) and (3), we obtain x^2-y^2=v^2-w^2 and solution for {v,w} is below.

w = -(-n^2t^2+m^2)/t, v = (n^2t^2+m^2)/t.

From equation (3), we obtain z^2 = (1-1/(t^2))m^4+(1-t^2)n^4........................(4)

Let U=m/n, we obain

V^2 = (1-1/(t^2))U^4+1-t^2..........................................................(5)

Equation (5) has a solution (U,V)=Q(t, t^2-1).

Hence this quartic equation (５) is birationally equivalent to an elliptic curve below.

Y^2+4tYX+8(t^4-2t^2+1)Y/t = X^3+(-6+2t^2)X^2-4(t^6-3t^4+3t^2-1)X/(t^2)-8(-6t^6+12t^4-10t^2+3+t^8)/(t^2)

Transformation is given, 
U = (4t^4+2t^2X-16t^2+tY+12-2X)/Y
V = (-24+104t^2-88t^2X+144t^4X+8t^5Y+8t^10+4X+144t^6-176t^4-56t^8+t^4X^3-72Xt^6-8tY+12t^8X+6t^6X^2-24t^4X^2+18t^2X^2-t^2X^3)/(t^2Y^2)
X = (2t^2V-2t^4-2V+2+4t^3U-4tU)/(U^2-2tU+t^2)
Y = (4t^4V-12t^4+4t^4U^2+16t^3U-8t^2V-16t^2U^2+16t^2-16tU-4+12U^2+4V)/(U^3-3tU^2+3Ut^2-t^3).


The point corresponding to point Q is P(X,Y)=(6-2t^2, -8(t^2+1)/t).

This point P is of infinite order, and the multiples kP, k = 2, 3, ...give infinitely many points.

Hence we can obtain infinitely many parametric solutions for equation (5).

Consequently, a simultaneous equation {x^2-y^2=u^2, x^2-z^2=v^2, y^2-z^2=w^2} has infinitely many parametric solutions.

Q.E.D.


3.Examples

When k=2 we obtain a parametric solution,

x = (t^2+1)*(t^8+20*t^6-26*t^4+20*t^2+1)
y = (t-1)*(t+1)*(t^4-4*t^3+10*t^2-4*t+1)*(t^4+4*t^3+10*t^2+4*t+1)
z = (t-1)*(t+1)*(t^4-4*t^3-6*t^2-4*t+1)*(t^4+4*t^3-6*t^2+4*t+1)
u = 2*(t^4+6*t^2-3)*t*(3*t^4-6*t^2-1)
v = 2*t*(t^4-2*t^2+5)*(5*t^4-2*t^2+1)
w = 8*t*(t-1)*(t+1)*(t^2+1)*(t^2-2*t-1)*(t^2+2*t-1).
t is arbitrary.

Similarly, when k=3 we obtain,
 
x = (t^2+1)*(t^16+104*t^14-548*t^12+3032*t^10-4922*t^8+3032*t^6-548*t^4+104*t^2+1)*(t^8+20*t^6-26*t^4+20*t^2+1)
y = (t-1)*(t+1)*(t^12-4*t^11+46*t^10+108*t^9-17*t^8+88*t^7+388*t^6+88*t^5-17*t^4+108*t^3+46*t^2-4*t+1)
  *(t^12+4*t^11+46*t^10-108*t^9-17*t^8-88*t^7+388*t^6-88*t^5-17*t^4-108*t^3+46*t^2+4*t+1)
z = (t-1)*(t+1)*(t^4-4*t^3+10*t^2-4*t+1)*(t^4+4*t^3+10*t^2+4*t+1)*(t^8+16*t^7+20*t^6-16*t^5-26*t^4-16*t^3+20*t^2+16*t+1)
  *(t^8-16*t^7+20*t^6+16*t^5-26*t^4+16*t^3+20*t^2-16*t+1)
u = 2*(t^4-2*t^2+5)*(t^8+52*t^6-26*t^4-12*t^2+1)*t*(5*t^4-2*t^2+1)*(t^8-12*t^6-26*t^4+52*t^2+1)
v = 2*t*(t^12-22*t^10+235*t^8-228*t^6+39*t^4+26*t^2+13)*(13*t^12+26*t^10+39*t^8-228*t^6+235*t^4-22*t^2+1)
w = 8*t*(t-1)*(t+1)*(t^2+1)*(t^2-2*t-1)*(t^2+2*t-1)*(3*t^4-6*t^2-1)*(t^4+6*t^2-3)*(t^4+4*t^3-6*t^2+4*t+1)*(t^4-4*t^3-6*t^2-4*t+1).

      

4.Reference

[1].L.E. Dickson:History of theory of numbers, vol 2.