1.Introduction

It's showed that {7x^2-y^2=3s^2, 7y^2-x^2=3t^2} has infinitely many integer solutions on MathStackExchange website[1].

We show that {ax^2+by^2=cs^2, ay^2+bx^2=ct^2} has infinitely many integer solutions.
                 
                 
2.Theorem
     
Simultaneous diophantine equation {ax^2+by^2=cs^2, ay^2+bx^2=ct^2} has infinitely many integer solutions.

condition: a = 16m^2, b = -9m^2+n^2, c = a+b.
m,n are arbitrary.

 
Proof.

ax^2+by^2=cs^2..........................................................(1)

ay^2+bx^2=ct^2..........................................................(2)

Let c = a+b.

From equation (1) we obtain a parametric solution as follows.

(x,y,s)=( ((a+b)k^2+(-2a-2b)k+a-3b)/f, ((-a-b)k^2+(-2a-2b)k+3a-b)/f, ((-a-b)k^2+(2a-2b)k-a-b)/f   )
f=(-1+k)(1+k)(a+b).

Substitute above (x,y) to equation (2) and let a = 16m^2, b = -9m^2+n^2, then we obtain quartic equation (3).

V^2 = (49m^4+14m^2n^2+n^4)U^4+(700m^4+72m^2n^2-4n^4)U^3+(-2402m^4+228m^2n^2-2n^4)U^2
    +(-2100m^4-216m^2n^2+12n^4)U-386m^2n^2+9n^4+5049m^4.................(3)

Since quartic equation (3) has a rational point Q(U,V)=(3, 64mn), then this quartic equation is birationally equivalent to an elliptic curve below.

Y^2+24m(5m^2+3n^2)YX/n+(164864m^5n+30720m^3n^3+1024mn^5)Y = X^3+16(139m^4n^2+21m^2n^4+n^6-225m^6)X^2/(n^2)+(-802816m^6n^2-229376m^4n^4-16384m^2n^6)X-959709184m^10n^2-720896000m^8n^4-126353408m^6n^6-9175040m^4n^8-262144m^2n^10+2890137600m^12

U = (128mn^2X+284672m^5n^2+43008m^3n^4+2048mn^6-460800m^7+3Yn)/(Yn)
V = (-184968806400m^13n^4+16777216m^3n^14+587202560m^5n^12+8086618112m^7n^10+46137344000m^9n^8+61421387776m^11n^6+32768mn^12X+3072mn^8X^2+64mn^4X^3+2424832m^3n^10X-2049638400m^11Xn^2+38240256m^5n^8X+227934208m^7n^6X+323452928m^9n^4X-691200m^7X^2n^2+64512m^3n^6X^2+427008m^5n^4X^2-491520m^8Yn^3-2277376m^6n^5Y+294912m^4n^7Y-8192m^2n^9Y+27648000m^10Yn+1658880000m^13X)/(Y^2n^3) 
X = (128mnV-5632m^2n^2+7680m^4U-23040m^4+4608m^2n^2U)/(U^2-6U+9)
Y = (16384m^2n^3V-333824m^3n^4-724992m^5n^2U-387072m^5n^2+331776m^3n^4U+284672m^5n^2U^2+43008m^3n^4U^2+2048mn^6U^2-12288mn^6U+18432mn^6-460800m^7U^2+2764800m^7U-4147200m^7)/(nU^3-9nU^2+27nU-27n)

The point corresponding to point Q is P(X,Y)=(-16(139m^4n^2+21m^2n^4+n^6-225m^6)/n^2, 128m(60m^6n^2+278m^4n^4-36m^2n^6-3375m^8+n^8)/n^3).

Hence we get 2Q(U,V)=((3n^8-28m^2n^6+146m^4n^4+228m^6n^2+675m^8)/((-m+n)(m+n)(-15m^3+7m^2n-mn^2+n^3)(15m^3+7m^2n+mn^2+n^3)),
64(-3m+n)(3m+n)(n^14+5m^2n^12-83m^4n^10-271m^6n^8-269m^8n^6-6049m^10n^4+6175m^12n^2+16875m^14)nm/((-m+n)^2(m+n)^2(-15m^3+7m^2n-mn^2+n^3)^2(15m^3+7m^2n+mn^2+n^3)^2)  ).

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

Hence we can obtain infinitely many integer solutions for simultaneous equation {ax^2+by^2=cs^2, ay^2+bx^2=ct^2}.

Case : h=1

x = 4m^2

y = 3m^2+n^2

s = 5m^2-n^2

t = 4mn


Case : h=2

x = 4(5625m^14-1075m^12n^2+9509m^10n^4+2553m^8n^6-101m^6n^8-137m^4n^10+7m^2n^12+3n^14)m^2

y = (n^2+3m^2)(-n^14-m^2n^12+27m^4n^10+427m^6n^8+2173m^8n^6-8291m^10n^4+16425m^12n^2+5625m^14)

s = (-n^2+5m^2)(-n^14-m^2n^12+27m^4n^10-597m^6n^8-4995m^8n^6-7267m^10n^4-9175m^12n^2+5625m^14)

t = 4(n^14+5m^2n^12-83m^4n^10-271m^6n^8-269m^8n^6-6049m^10n^4+6175m^12n^2+16875m^14)mn
       
Q.E.D.


3.Reference

[1]: StackExchange: Solving simultaneous (non linear) integer equations (a bit like conics)