Written: Oct. 23, 1997

Cecil Rousseau, the problem editor of SIAM Review, has just told me that SIAM Review's problem section's days are numbered. I was shocked, but not surprised. I have already had a premonition of that, a few months ago, when I read in SIAM News that SIAM Review is about to get a `face lift' and a `new image'. I am always wary of such proposed improvements that contribute to the contemporary trend to sacrifice content in favor of fluff and `image'. It is regrettable that even scientists, and mathematicians to boot, have caught the image-obsession that has turned politicians into puppets in the hands of sleazy PR-professionals.

The most interesting parts of the American Mathematical Monthly and SIAM Review are their problem sections. Nobody reads the articles, but many readers go straight to the problem section. Almost as interesting as the problems are the solutions. One of my favorite books is Klamkin's collection of problems from SIAM Reviews. Even from the snobbish, prestige-hungry, point of view of the SIAM administration, there is justification for the problem section. For example, Mehta's famous integral (that turned out to follow from Selberg's once-dormant 1944 paper), made its first appearance as a problem in SIAM Review. Similar things can be said about Monthly problems, for example, Erdos's problem from 1946 that started Euclidean Ramsey theory, and the famous Busseman-Petit conjecture that was first raised as a Monthly problem.

But, most importantly, Problem sections turned many young people into mathematicians. It was the late Joe Gillis's `Gilyonot leMatematika' (that hopefully still exists today), and especially its problem section, that made me, and many of my friends in Israel, into mathematicians.

I also feel a personal loss. My first `publication', in 1970, when I was an undergrad, was a solved problem in SIAM Review. Both the problem and my solution were real gems. I'll forget my right hand before I'll forget it. Let G be a finite group with n elements, and let S be any subset. Prove that S^n is a subgroup. My solution went as follows. When |S|=1 it is trivial. Otherwise, by pigeon-hole, there must be an i such that |S^(i)|=|S^(i+1)|, hence S^(i+1)=aS^i, for some a in G, and hence S^n=a^(n-i)S^i, S^(2n)=a^(2n-i)S^i=a^n S^n=S^n, hence (S^n)^2=S^n and S^n is a subgroup. I was so proud and delighted when I first solved it, and was really ecstatic when Murray Klamkin decided to publish my solution. I am sure that many had a similar experience.

In conclusion, let me quote Herb Wilf, who in a recent bio in the Monthly wrote (AMM 104 no. 6 (June-July 1997), p. 588): `Herbert Wilf has been editor of this Monthly, and remembers well how the `Problems and Solutions' tail often wagged the Monthly dog'. The same is true of SIAM Review, and it is a very stupid dog-owner that chops his dog's tail, especially one that wags so well.

Responses (and some factual corrections by Dick Askey)

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