does a slight change in a real value cause a massive change from finite to infinite?

Can anybody give an example of a system of equivalence and/or inequality relations dependent to just a real value number like $$\kappa$$, which satisfies for finitely many integers if $$\kappa =1$$ and holds for infinitely many integers if we change $$\kappa$$ from $$1$$ by infinitely small real value like $$\epsilon \in \mathbb{R}$$.

an example like the statement of ABC conjecture below, but surely be much simpler to be resolved in an easy way.

If :

i) $$\mathrm{rad}\,(n)$$ is the product of the distinct primes in $$n$$,

ii) $$A,B,C$$ are three positive coprime integers,

iii) $$A+B=C\$$,

iv) $$\kappa \in \mathbb{R}$$,

then for infinitely many exceptions we have: $$C<\mathrm{rad}\,(ABC)^{\kappa }\\ for\\ \kappa=1 .\tag{1}$$ and if we change $$\kappa$$ infinitely small $$\epsilon >0$$ from $$1$$ upward then for finitely many exceptions we have: $$C<\mathrm{rad}\,(ABC)^{\kappa }\\ for\\ \kappa=1+ \epsilon .\tag{2}$$

I asked this question because it is somehow odd to me that a slight change in a real value parameter may cause a sudden change to the state of a system dependent to that value from $$\aleph _0$$ to finite!

• Isn't your ABC example backwards? It has infinitely many exceptions if $\kappa=1$, but your question asks for a property that is satisfied for finitely many integers if $\kappa=1$. – TonyK Apr 7 at 23:01
• @TonyK does not change the generality of the question, it is enough to reverse the inequality mark and substitute infinite by finite. – MasM Apr 7 at 23:44

Let $$x$$ be the set of numbers $$x\in\mathbb{R}$$ satisfying $$1. Then for $$\kappa=1$$ there are no solutions, but for any $$\kappa>1$$ there are continuously infinite solutions!

If you want a set of solutions that are restricted to integers, consider the set of tuples $$(x,n)\in\mathbb{Z}^2$$ with $$1.

Edit: for a system where the number of solutions is finite but nonzero at $$\kappa = 1$$, take the following:

$$x\in\mathbb{Z}$$ such that $$1\leqslant x\leqslant\kappa^x$$.

Then when $$\kappa = 1$$, there is exactly one solution, namely $$x=1$$. But if $$\kappa > 1$$, then for large enough $$x$$ it will always be true that $$x<\kappa^x$$, giving infinitely many solutions.

• I'd bet that "finitely many solutions" was intended to exclude zero solutions, even though it wasn't explicitly stated. – JonathanZ supports MonicaC Apr 7 at 23:30
• Well done, can you give another example which instead of null solution for $\kappa=1$ in your second example ,there be finitely many solutions? – MasM Apr 7 at 23:35
• @MasM edited accordingly – Ethan MacBrough Apr 7 at 23:43
• Great, that implicitly implies how vast and mighty the real realm is, compare to countable world ;) – MasM Apr 8 at 1:17