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In the extant books of Diophantus, are considered in the system of equations. Of interest is the non-linear system of Diophantine equations. Some simple systems from his book manages to solve it.

For example, from 3 books tasks 10, 11. 4 books task 19. The solution presented in this. http://www.artofproblemsolving.com/community/c3046h1057324_the_system_is_almost_linear_diophantine_equations

But here's an example of a decision like almost linear system of 3 task book 6. Gives another view of the representation of solutions. http://www.artofproblemsolving.com/community/c3046h1055253_the_system_of_equations_15

But there is in the books such a group of tasks which are of the same type. The conditions are very similar. It is conceivable that they can be solved one way.

For example with 2 books tasks 20, 21. This system was considered Sierpinski. He found a private decision. And in the book of Diophantus referred to the decision according to his formulas don't work. It turned out that the formulas of the solutions may be several. http://www.artofproblemsolving.com/community/c3046h1046718__4

Now interested in the question itself.

In the 2nd book is very much the same type of systems that can be described as follows. I wrote to look for solutions in integers.

$$\left\{\begin{aligned}&X^2\pm{(X+Y)q}=Z^2\\&Y^2\pm{(X+Y)q}=R^2\end{aligned}\right.$$

$$\left\{\begin{aligned}&XY+(X+Y)q=Z^2\\&XY-(X+Y)q=R^2\end{aligned}\right.$$

$$\left\{\begin{aligned}&(X+Y)^2\pm{Xq}=Z^2\\&(X+Y)^2\pm{Yq}=R^2\end{aligned}\right.$$

The system is very similar and right of common and more simple approach to their solution. Because then there are systems with a large number of equations.

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The system of book 2 tasks 22 , 23 .

$$\left\{\begin{aligned}&X^2+(X+Y)q=Z^2\\&Y^2+(X+Y)q=R^2\end{aligned}\right.$$

Found this solution, but it sets a very different kind of decision. Different from the previous one.

$$X=9(p-s)(t-4p)$$

$$Y=9(s-p)(t-4s)$$

$$q=-64p^2+160ps-64s^2+2t^2-8t(p+s)$$

$$Z=(p-s)(60p-48s-3t)$$

$$R=(p-s)(48p-60s+3t)$$

$$***$$

$$X=s(pk^2+2skt-pt^2)$$

$$Y=s((p-2s)t^2-pk^2)$$

$$q=2pk((p-2s)t-pk)$$

$$Z=s(pk^2-2(p-s)kt+pt^2)$$

$$R=s(pk^2-2pkt+(p-2s)t^2)$$

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For the system of book 2 task 30.

$$\left\{\begin{aligned}&XY+(X+Y)q=Z^2\\&XY-(X+Y)q=R^2\end{aligned}\right.$$

Decisions will be.

$$X=(p^2+s^2)(p^2+s^2+2k^2)$$

$$Y=2k^2(p^2+s^2+2k^2)$$

$$q=2k^2(s^2-p^2)$$

$$Z=2ks(p^2+s^2+2k^2)$$

$$R=2kp(p^2+s^2+2k^2)$$

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The system of book 2: objectives 24, 25.

$$\left\{\begin{aligned}&(X+Y)^2+Xq=Z^2\\&(X+Y)^2+Yq=R^2\end{aligned}\right.$$

The solutions can be written as.

$$X=t^2+2(2p-s)t+3p^2-4ps+s^2$$

$$Y=2p(p+s-t)$$

$$q=8(s-p-t)(p+s-t)$$

$$Z=3t^2+6(p-s)t-p^2-6ps+3s^2$$

$$R=-t^2+2(3p+s)t+3p^2-6ps-s^2$$

Or so.

$$X=k^2(2kp^2-2kps+ts^2)$$

$$Y=k^2((t-2k)p+2ks)p$$

$$q=t^2(-2kp+(t-2k)s)s$$

$$Z=kt(kp^2-2kps+(t-k)s^2)$$

$$R=kt(-kp^2+(t-2k)ps+ks^2)$$

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2 books of Diophantus task $34 , 35$. For the system of Diophantine equations.

$$\left\{\begin{aligned}&a^2+(a+b+c)q=x^2\\&b^2+(a+b+c)q=y^2\\&c^2+(a+b+c)q=z^2\end{aligned}\right.$$

You can record such a decision.

$$a=nlt^2p^2+(n^2t^2+2nlk^2-k^2l^2)ps-nlk^2s^2$$

$$b=ktl^2p^2+((2k-t)tn^2+l^2k^2)ps-ktn^2s^2$$

$$c=((2k-t)n-kl)tlp^2+kn(kl+nt)s^2$$

$$q=2tls(kns+(kl-(2k-t)n)p)$$

$$x=nlt^2p^2+(t^2n^2-2k(k-t)nl+k^2l^2)ps+nlk^2s^2$$

$$y=ktl^2p^2+(k^2l^2+2ktnl-(2k-t)tn^2)ps+ktn^2s^2$$

$$z=(kl-(2k-t)n)tlp^2+2ktnlps+kn(kl+nt)s^2$$

$n,l,k,t,p,s - $ any integers.

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