## Diophantine equation Madame Zarangesh

Are there any ideas on solving niubd sleduyushego Diophantine equation? $\displaystyle X^3+Y^3+T^2=Z^4$

Blocked me and I can not answer the questions. Therefore, first edit. I'm talking about, this equation has already told, but this question asked on one site, well here and decided to redraw a more convenient form. solutions Diophantine equation Madame Zarangesh. $\displaystyle X^3+Y^3+T^2=Z^4$

You can record almost that trivial, but interesting is another case. When making formula takes some sort of a beautiful view.

I only solutions are obtained using solutions of Pell: For example, if we use the solutions: $\displaystyle p^2-3s^2=1$

Solutions have the form:

$\displaystyle Z=2p^2$

$\displaystyle X=2p(p+s)$

$\displaystyle Y=2p(p-s)$

$\displaystyle T=4p^2(p^2-1)$

If we make the change:

$\displaystyle a=2ps\pm(p^2+3s^2)$

$\displaystyle b=6ps\pm(p^2+3s^2)$

Then:

$\displaystyle Z=b^2$

$\displaystyle X=-b(b+a)$

$\displaystyle Y=-b(b-a)$

$\displaystyle T=b^2(b^2+2)$

If we use the solutions of Pell's equation: $\displaystyle p^2-15s^2=1$

And we use the substitution:

$\displaystyle a=6ps\pm(p^2+15s^2)$

$\displaystyle b=10ps\pm(p^2+15s^2)$

Then the solutions are of the form:

$\displaystyle Z=9a^2$

$\displaystyle Y=9a(a+b)$

$\displaystyle X=9a(a-b)$

$\displaystyle T=9a^2(9a^2-6)$

If the ratio of Pell's equation: $\displaystyle p^2-qs^2=1$

has the form $\displaystyle q=(k+1)^2+4(k^2-k+1)(2k^2-4k+1)$

$\displaystyle k$ - what some integer number of any character.

We make the change:

$\displaystyle a=(4k^2-6k+2)ps\pm(p^2+qs^2)$

$\displaystyle b=(8k^2-14k+6)ps\pm(p^2+qs^2)$

Then the solution can be written:

$\displaystyle Z=(k-1)^2(2a-(k+1)b)a$

$\displaystyle X=(k-1)^2(2a-(k+1)b)(a-b)$

$\displaystyle Y=(k-1)^2(2a-(k+1)b)(a-kb)$

$\displaystyle T=(k-1)^4(2a-(k+1)b)^2(a^2-1)$

Besides multiple solutions - when you come to solve the equations still to Pell's equation. It would be very interesting if you have an idea where it was possible to do without them. After substituting numbers for all primitive solutions will have to be divided into common divisor. Hopefully no need to explain? Although you can further complicate the formula, but I think it is not necessary. The meaning remains the same. I wonder whether the formula can look different?