The digital root of a prime number having the form m³-n³ is 1 or 7, where each of m and n is a positive integer, with m > n.

Is this assertion always valid?

If so, prove it.
If not, provide a counterexample.

It must be so.

Let x=m^3-n^3. Since x has to be prime, m and n must differ by 1, else (m-n) (m^2+m n+n^2) has at least 2 factors. But if so, then  (m+1)^3-m^3 ; 3m^2+3m+1. Ignoring primality for the moment, this produces x={1,7,19,37,61,91,...}. Now, from http://mathworld.wolfram.com/DigitalRoot.html, we need simply compute mod(x,9) in Excel, say, to confirm that for both prime and non-prime x the digital root of x is either 1 or 7, with period 3 {1,7,1,}.

The following generalisation is possible. 

Simplify 3(m^2+m)+1 to 3k+1; an expression which is again periodic, mod 9, with period 3 {4,7,1} so the initial problem resolves to the simpler one of dealing with the unwanted 4s generated by the 3k+1 expression.

It is clear that 3k+1 will be a difference of consecutive cubes iff k is twice a triangular number, since 3(n^2+n)+1 = 3n^2+3n+1. But no number of the form k=(3n+1) is twice a triangular number, 2T(n), as every such number 2T(n) is either divisible by 3, or leaves a remainder of 2 when divided by 3.

Hence every number of the form 3k+1, where k is not itself of the form (3n+1) has a digital root of either 1 or 7; and the problem is just a special example of this general principle.

Edited on May 4, 2014, 11:07 pm

Posted by broll on 2014-05-04 22:50:11