... from:

"PDEtools:-Solve(my_sol_1,y(x)); map(X->odetest(X,[ode,IC]),[%]) # [[0, 1], [0, undefined]]"

Maple points out that there is a problem with satisfying IC for the two explicit solutions. The theory of ordinary differential equations that I know only deals with equations in explicit form and, since the 1960s, also with so-called semi-implicit equations and ensures the existence and uniqueness of the solution. For general implicit equations, individual case investigations are therefore always necessary. This is obvious in the current case.

@dharr 

...for me, as a Maple beginner. In particular, solving the nonlinear system will keep me busy in the future.

Translation correction:

The initially confusing/complex Diophantine equation dio intrigued me because, after transforming the lhs into a sum of three squares, which equals zero, each term must be zero. This makes the solution obvious, or rather, results in three simple equations. This trick is well-known from competitions and always worth a try.

@dharr 

...for pointing out the * for commutative multiplication instead of . for non-commutative multiplication. I hadn't noticed that before and just typed it out of habit. Now expand works.

...sorry, I made a mistake when transferring the formula from the paper to Maple. It should be -1440*x*y^2 (the zero was missing).

@vv 

... "A sufficient condition for F(x,y)  to be Lipschitz wrt  y near (x0,y0)  is that D[2](F) be without singularities near (x0,y0). " in the current case, the classic proof: abs(f/x,u)-f(x,v))<=L*abs(u-v) is locally valid replaced. With which constant L should the proof be carried out locally and globally? And that is what my original question refers to.

@dharr 

Please excuse my mistake. I promise to do better.

...for both advices (dharr 8722, janhardo 750). Now I have a lot to practice, understanding how to apply the theory I know "from paper" to the maple world.

@dharr 

...I found my mistake.

@dharr @janhardo 

Several tests have shown that a,...,f = -17, -3, -7, 371, 113, 131 is causing problems. The solution should be (21; -5). I would appreciate some advice.

@dharr @janhardo 

I got identical results for both problems :-) . Their solution methods are clear thanks to good structuring and the presentation of relationships. This should be particularly valuable for students.

In contrast, the solution algorithm I know from other software works "hard" in number theory. Naturally, the discriminant is the starting point. A subsequent transformation (Legendre) yields the constant term, and then its factorization on the right-hand side of the equation takes place. And now, in most cases, the Chinese remainder theorem (Maple recognizes it) is applied modulo the prime factors, followed by an inverse transformation.

@janhardo 

...correct the input error, it was c = -17.

@janhardo 

Let the coefficients a, b, ..., f = 14, -5, -17, 2, 11, -12 and 3, -5, 7, 2, -11, 19. What does your algorithm say? Will the result include not only natural numbers but also positive and negative integers? Problems involving Diophantine quadratics in two variables are very interesting :-) . I have already performed calculations using external software and would like to compare them.

@janhardo 

...Countably many further solution pairs. These are to be calculated starting from Your basic solution (1,0) and (-1,0) using recursion. In this case, this is based solely on number theory and can be calculated using a third-party program (not Mathematica) based on it.

(BTW: I don't know if and which "names" can be mentioned here for legal reasons.)

@janhardo 

I don't know of a generally valid solution method for problems of this type. For this problem, you just have to come up with the right idea. I'll present it below in a somewhat simplified and abbreviated form.
So let n be the number you're looking for in the decimal system. The number n+1 is its successor. How do the digit sums Q(n) and Q(n+1) behave when n is converted to the number n+1? To do this, you first start by searching for possible/allowed numbers/digits, in the decimal representation of n, starting with the ones place at the far right end of the number. If there is a number n with the required property, then the digits 0,...,8 are eliminated in the ones place. This is because Q(n) and Q(n+1) would then be adjacent numbers, both of which cannot be divisible by 5. A chance of a solution only exists if the digit 9 is in the ones place.
It is clear that when calculating the cross sum Q, only the single-digit numbers/digits of the decimal representation of n play a role as summands. The power of ten, which describes the place value in the decimal system, is irrelevant here in the cross sum.
This process of elimination is now continued in the decimal representation of n, proceeding from right to left. If a solution is to be found, then all decimal places except the leftmost one must be filled with 9. For the cross sum Q(n), this leads to the suitable approach Q(n)=a+k*9 (k=1, 2, ...). Thus, one runs k starting from 1 and determines the smallest a that makes Q(n) divisible by 5, and then tests Q(n+1) for divisibility by 5. And for k=4 and a=4, the result is n=49999, with Q(n)=40 and Q(n+1)=5. This search algorithm actually found the smallest n with the required property.

@vv 

The AI ​​solution already assumes that "half" of the result is known... Then the solution is not a problem.
The real difficulty lies in recognizing this. Did I misunderstand the AI ​​solution?