# To solve the system of Diophantine equations.

I decided to compile a single task and to record such a system.

\left\{\begin{aligned}&xt+yw=az^2\\&xw-yt=br^2\end{aligned}\right.

$a,b -$ integers that are the problem. It is clear that it is necessary to solve the problem to get the formula. For $x,y,t,w,z,r -$ which is described in integers.

And at the same time to find out all the possible factors $a,b$ there are solutions?

It is clear that if to square one and another equation and then folded. You can get the product of the sum of squares, but further calculations will be numerical. A more interesting case when it is possible to write the formula itself.

For the system of equations:

\left\{\begin{aligned}&kx+ty=az^2\\&tx-ky=bq^2\end{aligned}\right.

Rewrite all the same this solution to compare.

$$k=2ps$$

$$x=2apsn^2+b(p^2-s^2)j^2$$

$$t=p^2-s^2$$

$$y=a(p^2-s^2)n^2-2bpsj^2$$

$$z=n(p^2+s^2)$$

$$q=j(p^2+s^2)$$

$a,b$ - coefficients specified by the problem statement. I was interested in other solutions when solutions are not multiples. Managed to get while this decision.

$$k=2ap^2-2bj^2+(4bj-(b+a)s)s$$

$$x=4a^2p^4-2a(4bj^2-4bjs-(b-a)s^2)p^2+4abps^3+4b^2j^4-$$

$$-8b^2sj^3+2b(b+a)s^2j^2+4b^2js^3-2b^2s^4$$

$$t=2ap^2+4aps+2bj^2+(a-b)s^2$$

$$y=4a^2p^4+8a^2sp^3-2a(4bj^2-8bjs+(3b-a)s^2)p^2-4ab(2j^2-4js+s^2)ps+$$

$$+4b^2j^4-16b^2sj^3+2b(11b+a)s^2j^2-12b^2js^3+2b^2s^4$$

$$z=4ap^3+4asp^2-2(2bj^2-4bjs+(b-a)s^2)p+4bjs^2-2bs^3$$

$$q=4a(s-j)p^2+4aps^2+4bj^3-8bsj^2+2(3b+a)js^2-2bs^3$$

$p,s,j,n$ - any integer asked us.