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.

Your Answer

 

By clicking "Post Your Answer", you acknowledge that you have read our updated terms of service, privacy policy and cookie policy, and that your continued use of the website is subject to these policies.

Not the answer you're looking for? Browse other questions tagged or ask your own question.