**POLYA010: Integers of the form bc/a + ca/b + ab/c
**(proposed
by Floor van Lamoen, 11/14/01).

For which positive integers n we can find three positive integers a,b,c
so that n = ab/c + ac/b + bc/a ?

**Discussion. [FvL, 11/16/01]: **Let me start discussion
myself. From bc/a + ca/b + ab/c = n we find a^2(b^2+c^2) + (bc-an)bc =
0, from

which we conclude that a|bc.

Now suppose gcd(a,b,c)=1. Then we can find P,Q,R,S with
gcd(P,Q)=gcd(R,Q)=gcd(S,P)=1 such that a=PQ, b=PR and c=QS. Since ac/b
is an integer, so is Q^2S/R, and thus R|S. In the same way from ab/c we
find S|R. So R=S, and n=P^2 + Q^2 + R^2. If we allow gcd(a,b,c)=T, then
simply n=T(P^2 + Q^2 + R^2).

So n = bc/a + ca/b + ab/c for positive integers a,b,c
is equivalent to n = d(a^2 + b^2 + c^2) for positive integers a,b,c,d.

**Bibliography. [1] POLYA003:**
Integers of the form a/(bc) + b/(ca) + c/(ab), proposed by J.P. Ehrmann.

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