This question already has an answer here:

I distinctly remember from my long-ago undergraduate math that there were some interesting cases where a solution (proof) was sought for some specific thing but it wasn't easy to find - and in a few of these cases a generalization was made to some wider problem and that turned out to be provable (maybe not easily, but understandably) and of course that in turn solved the specific case.

But I don't remember any such situation and would like to know of one or more.

(I don't think the whole issue of the power of complex numbers - e.g., in the Fundamental Theorem of Algebra or in complex analysis vs real analysis is what I'm thinking of. I'm sure there were more closely focused cases of this phenomenon that I once knew.)


marked as duplicate by Wojowu, Harry Gindi, Timothy Chow, Chris Godsil, YCor Feb 14 at 17:29

This question has been asked before and already has an answer. If those answers do not fully address your question, please ask a new question.

  • 2
    $\begingroup$ mathoverflow.net/questions/40005/… $\endgroup$ – Wojowu Feb 14 at 15:41
  • 2
    $\begingroup$ @HarryGindi, your duplicate comment is a duplicate of @‍Wojowu's duplicate comment. $\endgroup$ – LSpice Feb 14 at 16:06
  • 5
    $\begingroup$ @LSpice It gets added automatically when you vote to close with the reason being duplicate. $\endgroup$ – Harry Gindi Feb 14 at 16:11
  • $\begingroup$ Harry's correct. I have first voted to close on the basis of this being better fit on Math.SE, but later I have noticed this is a duplicate of a question on MO. This is why we've got the repeated comment. $\endgroup$ – Wojowu Feb 14 at 16:35
  • $\begingroup$ Ah, thanks for the pointers to the duplicate, I didn't find that. $\endgroup$ – davidbak Feb 14 at 18:09

Here's a very common type of example. Suppose you are interested in a certain convergent series $\sum_{n=0}^\infty a_n$. You look instead at the more general series $\sum_{n=0}^\infty a_n z^n$, to which you can apply the machinery of calculus, differential equations etc., and then specialize your result to $z=1$.

  1. There are many examples where this happens with properties concerning integers, where a property for a given integer may not be easy to prove, but the proof for the $n=0$ or $1$ case is easy, and induction is easy as well.

Alain Connes gave a great example of this on the radio you could even explain to a non-mathematician (actually that was the point): if I give you a chocolate bar with $8\times 5$ squares, and you're allowed to cut it by breaking any piece into two pieces along a straight line (not intersecting the squares), what is the way of doing it that requires the least amount of cutting ? For $8\times 5$ it's not clear, but if you generalize to $n\times m$ and use induction, then it becomes quite clear.

Perhaps a more mathematical example of this phenomenon would be quite appreciated: many divisbility questions fit into this, e.g. prove that $2^{2n}+2$ is divisible by $3$

  1. A different family of examples comes from applications of group theory or ring theory to number theory; for instance Euler's theorem : $a^{\varphi(n)} = 1 \mod n$ if $a\land n = 1$. I have never seen Euler's original proof of this, but seeing how complicated non-group-theoretic proofs of Fermat's little theorem are, I can only imagine that proving this only by number theory must be hard; while Lagrange's theorem is really easy and deals with this immediately if you know Bezout's theorem.

Another example in this family could be the fact that $n\mid \varphi(a^n-1)$; or $n! \mid \displaystyle\prod_{i=0}^{n-1}(q^n-q^i)$ whenever $q$ is a prime power (though for these, unlike with the first family of examples, it may not be clear how to go from the specific instance to the generalization)

  • $\begingroup$ The first example is terrific, thank you! $\endgroup$ – davidbak Feb 14 at 18:10

Not the answer you're looking for? Browse other questions tagged or ask your own question.