Answer to Question #149821 in Math for Wesam

Question

For any positive intege rx, let f(x) =x^x. Suppose that n is a positive integer such that there exists a positive integer m with m not equal 1 such that f(f(f(m))) =m^m^n+2020. Compute the smallest possible value of n.

Answer 1

$f(f(f(m))) = f(f(m^m))=f((m^m)^{m^m})=f(m^{m^{m+1}})=(m^{m^{m+1}})^{(m^{m^{m+1}})}=m^{m^{1+m^{m+2}}}$

Then find n by substituting it in equation:

$f(f(f(m)))=m^{m^n}+2020 \\ m^{m^{1+m^{m+2}}}=m^{m^n}+2020$

$n = log_m{[log_m{(m^{m^{1+m^{m+2}}}-2020)}]}$ -- is the strongly increased function and

$m^{m+2}<n<1+m^{m+2}$

Hence, the smallest possible value of n could be when m = 2: 8<=n<=9 and then $m^n\ge2^8>2$

But from other hand

$m^{m^{1+m^{m+2}}}=m^{m^n}+2020 \\ m^{m^{1+m^{m+2}}}-m^{m^n}=2^2*5*101\\ m^{m^n}*(m^{m^n(m^{1+m^{m+2}-n}-1)}-1)=2^2*5*101$

The left side number has prime factors with power more than 2: $m^n>2^8>2$ - contradiction.

Hence, n is not exist