A student I’m tutoring was working unsuccessfully on the following problem from the 2011 Mandelbrot Competition:
Let be positive integers such that . Find the minimum value of .
After some tinkering, I concluded that the problem as stated has no solution. I am now almost certain it was printed incorrectly: should be replaced by , and then we can solve the problem as follows:
It follows that . Since are positive integers we must have , and then it follows that the smallest solution occurs when . But what I’d like to discuss, briefly, is the argument showing that the misprinted problem has no solution.
Transcendental number theory
A general problem of the above type would appear at first glance to be quite difficult. However, conditional on Schanuel’s conjecture, these problems are in principle relatively straightforward.
Schanuel’s conjecture asserts that if are complex numbers which are linearly independent over , then
has transcendence degree at least over . It generalizes, among other things, both the Lindemann-Weierstrass theorem (the case where the are all algebraic) and the Gelfond-Schneider theorem (the case where for algebraic and algebraic irrational), two basic tools for proving transcendentality of various numbers.
For our purposes we want to specialize Schanuel’s conjecture as follows. If are distinct primes, then by unique factorization are linearly independent over , so it follows from Schanuel’s conjecture that
has transcendence degree exactly ; in other words, that the logarithms of the primes are algebraically independent. Even this special case of Schanuel’s conjecture is, to my knowledge, wide open. It follows in particular that the ring
is a polynomial ring on .
The misprinted problem
After writing everything in terms of natural logarithms of primes and clearing denominators, the misprinted problem is to find positive integers such that the polynomial
in is equal to the polynomial
But the polynomial ring in countably many variables is a UFD and are both linear, hence irreducible, so the part of the RHS not containing is necessarily divisible by either or , and it is straightforward to verify that neither of these holds. Thus there are no such .
The correct problem
Especially since the term already appears in the problem, the likeliest typo was that actually meant . In that case, after writing everything in terms of natural logarithms of primes and clearing denominators, the correct problem is to find positive integers such that
This time the RHS is divisible by , and dividing out gives as before.