I'm looking for problems that due to modern developments in mathematics would nowadays be reduced to a rote computation or at least an exercise in a textbook, but that past mathematicians (even famous and great ones such as Gauss or Riemann) would've had a difficult time with.

Some examples that come to mind are group testing problems, which would be difficult to solve without a notion of error-correcting codes, and -- for even earlier mathematicians -- calculus questions such as calculating the area of some $n$-dimensional body.

The questions have to be understandable to older mathematicians and elementary in some sense. That is, past mathematicians should be able to appreciate them just as well as we can.


closed as too broad by user147263, Jonas Meyer, user642796 May 7 '15 at 3:10

Please edit the question to limit it to a specific problem with enough detail to identify an adequate answer. Avoid asking multiple distinct questions at once. See the How to Ask page for help clarifying this question. If this question can be reworded to fit the rules in the help center, please edit the question.

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    $\begingroup$ Solvability by radicals and Galois theory.. I think it's only in the late 1870's that Galois theory has been understood! $\endgroup$ – mich95 May 6 '15 at 12:46
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    $\begingroup$ Factoring 20-digit numbers. $\endgroup$ – Gerry Myerson May 6 '15 at 12:59
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    $\begingroup$ Computers are mathematics. $\endgroup$ – Gerry Myerson May 6 '15 at 13:02
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    $\begingroup$ I don't understand why the question was put on hold. Certainly it is broad, but each answer can be objective, short and self-contained (as seen below). How can I make it fit MSE's rules? $\endgroup$ – mich May 6 '15 at 14:02
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    $\begingroup$ "I'm looking for problems that due to modern developments in mathematics would nowadays be reduced to a rote computation or at least an exercise in a textbook, but that past mathematicians (even famous and great ones such as Gauss or Riemann) would've had a difficult time with." Past mathematicians would have had trouble factoring a 20-digit number. Nowadays, it's a rote computation. Why is there an argument here? $\endgroup$ – Gerry Myerson May 7 '15 at 0:58

I would say that computing the Fourier coefficients of a tamed function is a triviality today even at an engineering math 101 level.

Ph. Davis and R. Hersh tell the long and painful story of Fourier series. I quote from their book:

"Fourier didn't know Euler had already done this, so he did it over. And Fourier, like Bernoulli and Euler before him, overlooked the beautifully direct method of orthogonality [...]. Instead, he went through an incredible computation, that could serve as a classic example of physical insight leading to the right answer in spite of flagrantly wrong reasoning."

(Fifth Ch. "Fourier Analysis".)


That there exist transcendental numbers. This was fist shown by Liouville, who proved that Liouville's number: $$\sum_{i=0}^\infty10^{-i!}$$ is transcendental.

The "modern" proof would be due to Cantor:

There are countably many algebraic numbers and uncountably many reals. Therefore there exists a transcendental number.

Proving that Liouville's number is transcendental isn't so hard, but compared to the above it seems quite torturous.

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    $\begingroup$ The problem of squaring the circle was of immense interest to the ancient mathematicians, which we now know to be impossible due to the transcendence of $\pi$. That is not to say that the proof $\pi$ is transcendental is easy, but it is understood. $\endgroup$ – JMoravitz May 6 '15 at 13:03

This sum-of-squares theorem of Fermat may qualify as an example:

An odd prime $p$ is expressible as the sum of squares $x^2+y^2$ if and only if $p\equiv 1 \text{ mod } 4$.

You can read this Wikipedia article (as of the most recent update to this answer) to see the difference in mental effort in the original proof by Euler, as opposed to a modern treatment using the fact that the Gaussian integers are a Euclidean domain.

A dual example: I think Brouwer would be astonished and pleased to know that the Brouwer fixed point theorem can now be proven for the simplex (and, with more effort, for convex polytopes) with absolutely no knowledge of topology; just some affine geometry and combinatorial intuition to prove Sperner's Lemma, and basic analysis to translate to the continuous setting.

It's still not an "easy" proof but it is an example of a classical problem that we now can solve with considerably less machinery, instead of the above example, whose ease of proof can be chalked up to more machinery.

  • $\begingroup$ Is this really a modern treatment, though, or a modern simplification? The question being whether Dedekind's proof is "easier" and more natural from the perspective of modern theory, as opposed to just being a very clever simplification enabled by modern tools (but which would've required large mental effort to come up with). $\endgroup$ – mich May 6 '15 at 13:44
  • $\begingroup$ @Michael: Yeah, you can see some of the earlier edits where I was a bit more bold in my claim :P . I feel justified in saying that it's not "just" a simplification: Dedekind's proof, even after laying out all the details, is substantively different than Euler's. But there are definitely details which expand the length (and difficulty) of Dedekind's proof considerably. $\endgroup$ – Eric Stucky May 6 '15 at 13:49
  • $\begingroup$ What's a good resource to learn about this "Euclidean domain" thing? $\endgroup$ – Akiva Weinberger May 6 '15 at 18:37
  • $\begingroup$ Any good intro abstract algebra textbook should explain basic ring theory. $\endgroup$ – Eric Stucky May 9 '15 at 5:47

In the nineteenth century expressing the antiderivative of an elementary function as an elementary function was an open problem.

Nowadays, Risch algorithm, which can be run on machines, decides whether such operation can be done and, if so, yields a version of the correct result.

I cannot speak for past mathematicians, but I think this is a useful tool.

Added: @columbus8myhw made a very important technical remark in the comments, which is also explained in the last part of the wikipedia article.

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    $\begingroup$ Technically, the Risch algorithm can't fully be run on a computer — because it involves deciding whether a given equation is always equal to $0$, which is an unsolvable problem. (IIRC.) But there are heuristic algorithms, so it can be run well enough. $\endgroup$ – Akiva Weinberger May 6 '15 at 15:07

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