# Parametrization of solutions of diophantine equation

The issue I discussed in this thread. Parametrization of solutions of diophantine equation $x^2 + y^2 = z^2 + w^2$

Generally speaking at the forum often ask a question about this equation. So I think that will not solve different each time Diophantine equation is better to write the equation in this General form:

$$ax^2+bxy+cy^2=ez^2+jzw+tw^2$$

$a,b,c,e,j,t -$ integer coefficients which are defined by the problem statement.

The task is simple - to write a formula describing the parameterization of the equation. The formula itself and will specify conditions when possible integer solutions.

Many people like Diofantos geometry, but its methods are known for a very long time - here is inefficient. It is always better to have a single formula describing all equations than every time to solve the new equation.

• No need for the cross-terms. This can be easily transformed to, $$Ap^2+Bq^2 = Cr^2+Ds^2$$ – Tito Piezas III Jan 31 '15 at 13:39
• @TitoPiezasIII this form may not always be converted into such a form. – individ Jan 31 '15 at 13:42
• Actually, it is a well-known linear transformation. Let $x,\,y = p + m_1 q,\; n_1 q$, and $z,\,w = r + m_2 s,\; n_2 s$ to get $$a p^2 + (2 a m_1 + b n_1) p q + (a m_1^2 + b m_1 n_1 + c n_1^2) q^2 = \\e r^2 + (2 e m_2 + j n_2) r s + (e m_2^2 + j m_2 n_2 + t n_2^2) s^2$$ then choose $m_1,\,n_1,\,m_2,\,n_2$ such that, $$2 a m_1 + b n_1 = 0$$ $$2 e m_2 + j n_2 = 0$$ – Tito Piezas III Jan 31 '15 at 14:13
• @ТитоPiezasIII I said wrong. The formula for the solution in the General form contains all the factors that I wrote. And not always written in such a form can give solutions. When you write a formula then we'll see. – individ Jan 31 '15 at 15:31
• @ТитоPiezasIII The formula for this equation I have written there. artofproblemsolving.com/blog/98937 artofproblemsolving.com/blog/98917 artofproblemsolving.com/blog/98916 But it is necessary to record in another form of this solution. – individ Feb 1 '15 at 9:28

Equation if we write in the General form:

$$aX^2+bXY+cY^2=eZ^2+jZW+tW^2$$

If in this equation there any equivalent to a quadratic form in which the root is an integer.

$$q=\sqrt{b^2+4a(e+j+t-c)}$$

Then there are solutions. They can be written by making the replacement.

$$x=(b(2(e+j+t)-b)+4ac)s-(b+2a)(j+2t)k$$

$$y=(b^2+4c(e+j+t-a))s^2-2(b+2c)(j+2t)sk+(j^2+4t(a+b+c-e))k^2$$

Then decisions can be recorded and they are as follows:

$$X=(b-2(e+j+t-c)\pm{q})p^2+2(q((j+2t)k-(b+2c)s)\pm{x})pn+$$

$$+(((2(e+j+t-c)-b)\pm{q})y+2((j+2t)k-(b+2c)s)x)n^2$$

$$***$$

$$Y=(\pm{q}-(b+2a))p^2+2(q((j+2t)k-(2(e+j+t-a)-b)s)\pm{x})pn+$$

$$+(((b+2a)\pm{q})y+2((j+2t)k-(2(e+j+t-a)-b)s)x)n^2$$

$$***$$

$$Z=(\pm{q}-(b+2a))p^2+2(q((j+2t)k-(b+2c)s)\pm{x})pn+$$

$$+(((b+2a)\pm{q})y+2((j+2t)k-(b+2c)s)x)n^2$$

$$***$$

$$W=(\pm{q}-(b+2a))p^2+2(q((2(a+b+c-e)-j)k-(b+2c)s)\pm{x})pn+$$

$$+(((b+2a)\pm{q})y+2((2(a+b+c-e)-j)k-(b+2c)s)x)n^2$$

$p,n,k,s$ - integers asked us.