@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?

@vv 

...for this very instructive solution. It contains nested commands and a procedure with "for" and "if" statements. Back then, the solution involving paper and pen was very strenuous for me.

@janhardo 

This way of thinking and working is completely new to me. So far, I have solved all the tasks myself, down to the last detail. Thanks for the tip. What AI is being used?

@janhardo 

I can decipher the AI ​​procedure. I understand it. I consider deciphering it pointless. There is a much simpler solution, such as the one presented by @vv13957, and this was precisely my previous approach. It is generalizable. Since I'm not interested in solving the problem but rather in the elegance of a solution (and in this case, also in learning the Maple procedure technique), I don't want to go into it in more detail here. The explanations would require a lot of writing, which isn't necessary.
I'm very cautious about AI and sometimes take critical, even questionable, positions:
-AI isn't capable of creativity, but rather applies known knowledge exclusively in previously unknown combinations.
-AI isn't reliably capable of self-diagnosis (in the spirit of the logic classic "all Greeks are liars").

In any case, the procedural solutions found here hit the mark in my view – not only a solution, but also instructive.

@dharr 

...to put the term "birthday child" (regardless of age) in "...", which is common in my home country.

Both solutions (also @vv 13942) are instructive for me as an introduction to procedural technology.

@janhardo 

... the procedure from  vv 13937 helps me to get started with "for" and "if".

@janhardo 

...I was hoping for a Maple procedure as a solution ;-). I'd like to explore such procedures in more detail in the future.

... to your example 4 ;-) .
Determine all intervals on the x-axis where the Lipschitz constant 0<L<1.

The solution to the problem becomes obvious when the ODE is solved in explicit form.

@acer 

...for Your efforts and constructive help. It's especially unusual for me to learn that commands with assigned variables can themselves be used as variables in "outer" commands (nesting). Learned something new again :).

@acer 

Thanks for the help. I'm having trouble understanding the nested command "evalf(collect(simplify(evala(sol)), exp))". What role does "exp" play in the "collect" option?

@sand15 

...ow can the cumbersome numerical terms in the solution y(x) be converted to floating-point numbers?

@sand15 

... A direct hit, that's it.