Titu generalizes this to the case in which positive even integer k is in place of 1987. The number of such function is k!/(½k)!, so there are no such functions for odd k
f : ℕ₀→ℕ₀ where ℕ₀=ℕ∪{0} f(f(n))=n+k f(n+k)=f(f(f(n)))=f(n)+k thus, f(n+qk)=f(n)+qk for any integer q All is left is to check the values of f for n∈{0,1,...,k-1}=ℤₖ suppose t∈ℤₖ f(t)=qk+r, for some r∈ℤₖ t+k=f(qk+r)=qk+f(r) t-(q-1)k=f(r)≥0 (q-1)k≤t
Titu generalizes this to the case in which positive even integer k is in place of 1987. The number of such function is k!/(½k)!, so there are no such functions for odd k
Thank you for sharing this. This is very interesting.
f : ℕ₀→ℕ₀
where ℕ₀=ℕ∪{0}
f(f(n))=n+k
f(n+k)=f(f(f(n)))=f(n)+k
thus, f(n+qk)=f(n)+qk for any integer q
All is left is to check the values of f for n∈{0,1,...,k-1}=ℤₖ
suppose t∈ℤₖ
f(t)=qk+r, for some r∈ℤₖ
t+k=f(qk+r)=qk+f(r)
t-(q-1)k=f(r)≥0
(q-1)k≤t
Titu's 2007 book is so good
Great Video!!
thanks!!
Thank you so much
Can you make a video about P4 from the canadian math olympiad 2021
It was an elegant GCD functional equation
I will take a look!