1.Introduction

Is there a solution of simultaneous equation {a^2+b^2=x^2, a^2+c^2=y^2, b^2+c^2=z^2}? 

According to Dickson's book[1], L.Euler gave a parametric solution below.
a = 8f(f^4-1)
b = (1-f^2)(f^4-14f^2+1)
c = 2f(3f^4-10f^2+3)

Saunderson gave a parametric solution as follows.

a = a(4q^2-r^2)
b = b(4p^2-r^2)
c = 4pqr
p^2+q^2=r^2

We show simultaneous equation {a^2+b^2=x^2, a^2+c^2=y^2, b^2+c^2=z^2} has infinitely many parametric solutions.
     
2.Theorem
     
A simultaneous equation {a^2+b^2=x^2, a^2+c^2=y^2, b^2+c^2=z^2} has infinitely many parametric solutions.
 
Proof.

a^2+b^2=x^2...............................................................(1)

a^2+c^2=y^2...............................................................(2)

b^2+c^2=z^2...............................................................(3)

From equation (1) and (2) we obatian

a = 2pq, b = p^2-q^2, c = p^2q^2-1........................................(4)
x = p^2+q^2, y = 1+p^2q^2.................................................(5)


Substitute (4) and (5) to equation (3), we obtain a quartic eqution.

Let U = p and V = z.

V^2 = (q^4+1)U^4-4U^2q^2+q^4+1............................................(6)

Quartic equation (6) has a solution (U,V) = Q(q,q^4-1).

Hence this quartic equation (6) is birationally equivalent to an elliptic curve below.


Transformation is given, 

U = (4q^10+2q^4X-4q^2+qY-2X)/Y
V = (8q^6-8q^2-4X+8q^22-16q^14-X^3+16q^10-8q^18-8q^8X+8qY+6q^10X^2+12q^16X+q^4X^3-6X^2q^2-8q^9Y)/(Y^2)
  
X = (2q^4V-2q^8-2V+2-4q^3U+4q^7U)/(U^2-2Uq+q^2)

Y = (4q^10U^2+4q^8+4q^8V-16q^7U-8q^4V+16q^3U-4q^2U^2-4+4V)/(U^3-3U^2q+3Uq^2-q^3)

The point corresponding to point Q is P(X,Y) = (-2q^2-2q^6, 8q+8q^5).

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 (7).

Consequently, a simultaneous equation {a^2+b^2=x^2, a^2+c^2=y^2, b^2+c^2=z^2} has infinitely many parametric solutions.

Q.E.D.


3.Examples


When k=2 we obtain a parametric solution,

a = 2*q^2*(q^4-3)*(3*q^4-1)
b = 8*q^2*(q^8-1)
c = (q^4-1)*(q^4-4*q^2+1)*(q^4+4*q^2+1)
x = 2*q^2*(5*q^8-6*q^4+5)
y = (q^4+1)^3
z = (q^4-1)*(q^8+18*q^4+1)

p,q is arbitrary.

Similarly, when k=3 we obtain a parametric solution,

a = 2*(q^12-41*q^8+27*q^4+5)*q^2*(5*q^12+27*q^8-41*q^4+1)
b = 8*q^2*(q^8-1)*(q^4-3)*(3*q^4-1)*(q^8+18*q^4+1)
c = (q^4-1)*(q^4-4*q^2+1)*(q^4+4*q^2+1)*(q^8+8*q^6-14*q^4+8*q^2+1)*(q^4-4*q^3+4*q^2-4*q+1)*(q^4+4*q^3+4*q^2+4*q+1)
x = 2*q^2*(13*q^24+94*q^20+1027*q^16-2204*q^12+1027*q^8+94*q^4+13)
y = (q^4+1)*(q^24-58*q^20+2063*q^16-3948*q^12+2063*q^8-58*q^4+1)
z = (q^4-1)*(q^24+182*q^20-561*q^16+2612*q^12-561*q^8+182*q^4+1)


4.References

[1].L.E. Dickson:History of theory of numbers, vol 2.
[2].Tito Piezas: Euler Bricks and Euler Quadruples.