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POLYA014:  Square as sum of two factorials (proposed by Lev Emelyanov, 11/18/01).

2^2=2!+2!; 5^2=4!+1!; 11^2=5!+1!; 12^2=5!+4!; 71^2=7!+1!. Does there exist any more n^2=k!+m!?
 

Discussion. [SCL, 11/26/01]: I have only a small contribution on this one, but perhaps it will give somebody else a clue.
Let f(q) denote the number of zeroes at the end of q, and let g(k)=f(k!). That is g(k)=n/5+n/25+n/125+..., where each quotient is truncated. (In TeX, this would look something like $g(k)=\lfloor n/5 \rfloor + \lfloor n/25 \rfloor + \lfloor n/125 \rfloor + \lfloor n/6255 \rfloor + ...$)
 Suppose that k\le m.  Then, if k!+m! is a square, f(k!+m!) is even. Now, if g(k) is odd, g(k)=g(m), limiting the possible values of m for such a k: m\le k+4- (k mod 5).
 Maybe somebody could do a bit more by looking at the last non-zero digit. For a square, this would be a 1, 4, 9, 6, 5, but for factorials (after the initial segment), it is 2, 4, 6, or 8. If k<m, then k! must end  ...4000...000 or ...6000...000, again limiting the k that need to be checked.
 I've done a little computer run without either of these tricks, and there are no other solutions besides the five shown in the statement of the problem, with 1\le k\le 560 and k\le m\le 1000. (That is, any other example either has k>560 or m>1000.)
Obviously, with the tricks, I could have saved about three quarters of the time of the computer run, since g(k) rates to be odd approximately half the time, and k! has last non-zero digit 4 or 6 about half the time (at a rough guess), and I assume these are somewhat independent.
 
 

Bibliography.

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