Functional Diophantine Equation

Number Theory Level 5

Let $f : \mathbb{N} \to \mathbb{N}$ be a strictly increasing function such that $f(2) = 8$ and $f(ab) = f(a) \cdot f(b)$ for $\gcd(a, b) = 1$ .

Evaluate the number of triples of positive integers $(a,b,n)$ satisfying the equation $f(n) = a^3 + b^3.$

The answer is 0.

1 solution

Ajinkya Shivashankar
Feb 4, 2017

Given that $f(ab) = f(a) \cdot f(b)$ . Thus, we can see that $f(x)=x^n$ is a possible solution. Also $f(2)=8$ so we get n=3 i.e $f(x)=x^3$

We need to find positive integers such that $f(n) = a^3 + b^3$ or $n^3 = a^3 + b^3$ .

By Fermat's Last theorem , this has no solutions. Thus answer is 0 .

To see why $f(x)=x^n$ is a solution to this, let $g(x)=\ln(f(x)$ Taking log on both sides of given relation, we get $g(ab)=g(a)+g(b)$ which has solution as $g(x)=k\ln(x)$ . Thus we get our required relation.

Ajinkya Shivashankar - 4 years, 4 months ago

Can you prove that it is the only solution?

Ishan Singh - 4 years, 4 months ago

There are a lot of solutions to $g(ab) = g(a) + g(b)$ that are not of the form $g(x) = k \ln (x)$ .

E.g. $g(2^k n) = k$ where $n$ is an odd number.

You need to use the condition of increasing function to work around it.

Calvin Lin Staff - 4 years, 4 months ago

Functional Diophantine Equation

The answer is 0.

1 solution

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