Prove that there are only finitely many cuboidal blocks with integer sides a x b x c, such that if the block is painted on the outside and then cut into unit cubes, exactly half the cubes have no face painted.
Solution
It is sufficient to show that abc = 2(a - 2)(b -2)(c - 2) has only finitely many solutions in integers c ≤ b ≤ a. If c ≤ 4, then c ≥ 2(c - 2), but ab > (a - 2)(b - 2), so there are no solutions. If c ≥ 10, then c/(c - 2) ≤ 10/8, so D ≤ 125/64 < 2, where for convenience we have written abc/( (a - 2)(b - 2)(c - 2) ) as D, so there are no solutions. Hence c = 5, 6, 7, 8, or 9.
If c = 5 and b ≥ 24, then D ≤ 5/3 (24/22)2 < 2, so there are no solutions. Now a/(a - 2) is strictly monotonic decreasing, so there is at most one solution for given b, c. Hence there are only finitely many solutions for c = 5.
Similarly, for c = 6, we find that are only solutions for b < 16; for c = 7, we find b < 14; for c = 8, b < 12 and for c = 9, b < 12. Hence there are only finitely many solutions in each case.
© John Scholes
jscholes@kalva.demon.co.uk
5 Mar 2002