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I'm looking for a method to find the parametric solutions of a Diophantine equation of this kind

$$ \alpha x^2 +\beta y^2 = \gamma w^2+\delta z^2 $$

$(\alpha,\beta,\gamma,\delta,x,y,w,z) \in \mathbb{N}$

es: $$ 3x^2+y^2=2w^2 +2z^2 $$ $$3x^2+2y^2=4w^2+z^2 $$

thank you

Patrick Danzi
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    https://math.stackexchange.com/questions/1127654/parametrization-of-solutions-of-diophantine-equation – individ Jul 03 '19 at 04:26
  • @individ thanks for the answer, I have some questions about it: can I obtain the solutions to my equation by setting b = j = 0? Since (± q, ± x) appears in the solutions, assigned (s, k, n, p, q, x, y), and chosen for example (-q, + x) in X, this choice of (-, +) should it also be used in Y, Z, W? assigned eg: (+, +) then {X (+, +); Y (+, +); Z (+, +); W (+, +)}, with the variation of (s, k, n, p) give me all the possible integer solutions for the equation? (Including those found starting from a different choice of (±, ±)?) thank you – Patrick Danzi Jul 05 '19 at 14:36
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    Yes. Certainly. If the formula does not provide a solution. Then you need to make a replacement. $X=x+ky$ We obtain an equation of equivalent form. And we look for such coefficients when the root is an integer. This usually comes to clarify the question of the solvability of some equivalent Pell equation. – individ Jul 05 '19 at 15:19
  • @individ but if the formula gives a solution then automatically i ve all the others? as coefficients to look for do you mean k? – Patrick Danzi Jul 05 '19 at 16:13
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    If the formula gives a solution. Then nothing else is required. If the root is intact. If the root is not whole. Then you need to look for the equivalent form of the equation in which the root is whole. – individ Jul 05 '19 at 17:08

5 Answers5

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In either case it is easy enough to find all rational solutions by stereographic projection around a known solution, both have (1,1,1,1). Take a relatively prime quadruple $(p,q,r,s)$ and evaluate the quadratic form at $$ (1,1,1,1) + t (p,q,r,s). $$ The first form is $3x^2 + y^2 - 2 z^2 - 2 w^2.$ The second is $3x^2 + 2 y^2 - z^2 - 4 w^2.$ In either case $t=0$ is always a solution, then there is a nonzero rational value for $t$ that gives a solution in that direction, and $t$ depends on the direction integers $(p,q,r,s).$

Give me a few minutes. Note that sorting out the primitive integral solutions is the real problem and harder.

Here is the calculation for the first using gp-pari. I will also typeset it. Before I do that, note that, even though we take $\gcd(p,q,r,s) = 1,$ it is still easy to get common factors in $(x,y,z,w).$ So, one half-satisfactory way to proceed is just to say "use these formulas and then divide out by any $\gcd(x,y,z,w).$" ADDED: I am not so sure the set of possible gcd's is finite with four variables. It is in three variables, goes back to Fricke and Klein (1897). I did a trial run with bounds on $p,q,r,s,$ the list of all gcds was:

   0   1   2   3   4   6   8   9  11  12  13  16  18  22  23  24  25  26  27  32  33  36  37  39  44  46  47  48  49  50  52  54  64  66  72  74  75  78  81  88  92  94  96  98 100 104 108 118 121 122 128 132 138 142 144 146 147 148 162 166 169 184 192 196 200 222 242 243 256 288 300 324 338 363 392 400 432 484 507 576 588 676 784

$$ x = -3p^2 + q^2 - 2 r^2 - 2 s^2 - 2 pq + 4pr + 4ps \; , \; $$ $$ y = 3p^2 - q^2 - 2 r^2 - 2 s^2 - 6 pq + 4qr + 4qs \; , \; $$ $$ z = 3p^2 + q^2 + 2 r^2 - 2 s^2 - 6 pr -2qr + 4rs \; , \; $$ $$ w = 3p^2 + q^2 - 2 r^2 + 2 s^2 - 6 ps -2qs + 4rs \; , \; $$

See how, if we simply take $p=1$ and the others $0,$ we get $(-3,3,3,3)$ with a common factor of $3$

parisize = 4000000, primelimit = 500000
? d = 3 * p^2 + q^2 - 2 * r^2 - 2 * s^2
%1 = 3*p^2 + (q^2 + (-2*r^2 - 2*s^2))
? n = -6 * p - 2 * q + 4 * r + 4 * s
%2 = -6*p + (-2*q + (4*r + 4*s))
? x = d + n * p
%3 = -3*p^2 + (-2*q + (4*r + 4*s))*p + (q^2 + (-2*r^2 - 2*s^2))
? y = d + n * q
%4 = 3*p^2 - 6*q*p + (-q^2 + (4*r + 4*s)*q + (-2*r^2 - 2*s^2))
? z = d + n * r
%5 = 3*p^2 - 6*r*p + (q^2 - 2*r*q + (2*r^2 + 4*s*r - 2*s^2))
? w = d + n * s
%6 = 3*p^2 - 6*s*p + (q^2 - 2*s*q + (-2*r^2 + 4*s*r + 2*s^2))
? 
? 
? 
? 3 * x^2 + y^2 - 2 * z^2 - 2 * w^2
%7 = 0
?
Will Jagy
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  • thanks for the answer, some things are not clear to me though: (p, q, r, s) must be a vector of prime numbers? your parameterization for (x, y, w, z) divided by gcd (x, y, w, z) gives all possible integer solutions for the equation I proposed? Moreover it is not very clear to me that the passage from the vector (1 + t p, 1 + t q, 1 + t r, 1 + t s) to the obtainment of the parameter solutions for x, y, w, z – Patrick Danzi Jul 05 '19 at 15:01
  • I watched this video and I understood: https://www.youtube.com/watch?v=UHDni7zhfOs. Thank you for your answer – Patrick Danzi Jul 10 '19 at 10:52
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Above equation shown below:

$3x^2+2y^2=z^2+4w^2$

Above has parameterization:

$x=k^2+6k+3$

$y=7k^2+4k-1$

$z=k^2-6k-5$

$w=5k^2+4k+1$

For, $k=7$, we get:

$(x,y,z,w)=(94,370,2,274)$

After removing common factor's we get:

$3(47)^2+2(185)^2=(1)^2+4(137)^2$

Sam
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As for the Pythagorean triples, the parametrization of some quadratic Diophantine equations comes from identities such as $(rX + sY) ^ 2 + (rY-sX) ^ 2 = (rX-sY) ^ 2 + (rY + sX) ^ 2$ that parameterizes the equation $x ^ 2 + y ^ 2 = z ^ 2 + w ^ 2$. But if this last equation goes with coefficients other than $1$, then there are no previously known identities that parameterize them. And calculations can be cumbersome without the help of specific programs in calculators. However, there is a theorem about the existence of solutions for quadratic equations .

Theorem.-$F(X)=ax^2+by^2+cz^2+dw^2=0$ is solvable in $\mathbb Q^*$ if and only if $F(X)\equiv0\pmod{p^r}$ is solvable for all primes $p$ and all power $r\in\mathbb N$ with $(x,y,z,w,p)=1$ and $F(X)=0$ is solvable in $\mathbb R$.

(note that, for example $F (X) = x ^ 2 + y ^ 2 + z ^ 2 + w ^ 2 = 0$ has no real solution not null).

Piquito
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"OP" requested solution for below mentioned equation:

$ax^2+by^2=cz^2+dw^2$ --------(1)

Let, $ap^2+bq^2=cr^2+ds^2$ ----$(2)$

From known numerical solution to equation $(2)$,

another numerical solution can be found from

the formula given below with the condition,

$[(a+b)=(c+d)]$,

The solution is:

$(x,y,z,w)=[(p+6w),(q+4w),(r+6w)(s+4w)]$

Where, $w$=$(3(cr-ap)+2(ds-bq))/(5(a-c))$

For, $(a,b,c,d)=(5,3,1,7)$ & $(p,q,r,s)=(13,3,23,7)$ we get,

after removing common factors:

$(x,y,z,w)=(4,31,46,11)$

$5(4)^2+3(31)^2=1(46)^2+7(11)^2$

Sam
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"OP" requested solution for the below mentioned equation,

$ax^2+by^2=cy^2+dw^2$ -------$(A)$

There is the Identity,

$(a+b)^3=a(a-3b)^2+b(3a-b)^2$ ------$(1)$

Also,

$(c+d)^3=c(c-3d)^2+d(3c-d)^2$ -----$(2)$

We take, $[(a+b)=(c+d)]$

Equating equations (1) & (2) we get:

$a(a-3b)^2+b(3a-b)^2=c(c-3d)^2+d(3c-d)^2$

Hence solution for equation "(A)" is:

$(x,y,z,w)=[(a-3b),(3a-b),(c-3d),(3c-d)]$

For, $(a,b,c,d)=(7,2,5,4)$ we get,

$7(1)^2+2(19)^2=5(7)^2+4(11)^2=(9)^3=(3)^6$

Sam
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