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Markus Scheuer
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Assume strictly monotone increasing function; such that $f:N^{+}\to N^{+}$, $h$ for all $n\in N^{+}$, $$f(f(f(n)))=f(f(n))\cdot f(n)\cdot n^{2015}$$

Prove or disprove:$f(n)=n^{13}$

Put $n=1,f(1)=m$ $$f(f(m))=mf(m)$$ Put $n=m$, $$f(f(f(m)))=f(f(m))f(m)m^{2015}\Longrightarrow f(mf(m))=m^{2016}(f(m))^2$$ What about following?

Assume strictly monotone increasing function; such that $f:N^{+}\to N^{+}$, $h$ for all $n\in N^{+}$, $$f(f(f(n)))=f(f(n))\cdot f(n)\cdot n^{2015}$$

Prove or disprove:$f(n)=n^{13}$

Put $n=1,f(1)=m$ $$f(f(m))=mf(m)$$ Put $n=m$, $$f(f(f(m)))=f(f(m))f(m)m^{2015}\Longrightarrow f(mf(m))=m^{2016}(f(m))^2$$ What about following?

Assume strictly monotone increasing function; such that $f:N^{+}\to N^{+}$, $h$ for all $n\in N^{+}$, $$f(f(f(n)))=f(f(n))\cdot f(n)\cdot n^{2015}$$

Prove or disprove:$f(n)=n^{13}$

Put $n=1,f(1)=m$ $$f(f(m))=mf(m)$$ Put $n=m$, $$f(f(f(m)))=f(f(m))f(m)m^{2015}\Longrightarrow f(mf(m))=m^{2016}(f(m))^2$$ What about following?

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Assume strictly monotone increasing function; such that $f:N^{+}\to N^{+}$, $h$ for all $n\in N^{+}$, $$f(f(f(n)))=f(f(n))\cdot f(n)\cdot n^{2015}$$

Prove or disprove:$f(n)=n^{13}$

Put $n=1,f(1)=m$ $$f(f(m))=mf(m)$$ Put $n=m$, $$f(f(f(m)))=f(f(m))f(m)m^{2015}\Longrightarrow f(mf(m))=m^{2016}(f(m))^2$$ What about following?

Assume strictly monotone increasing function; such that $f:N^{+}\to N^{+}$, $h$ for all $n\in N^{+}$, $$f(f(f(n)))=f(f(n))\cdot f(n)\cdot n^{2015}$$

Prove or disprove:$f(n)=n^{13}$

Put $n=1,f(1)=m$ $$f(f(m))=mf(m)$$ Put $n=m$, $$f(f(f(m)))=f(f(m))f(m)m^{2015}\Longrightarrow f(mf(m))=m^{2016}(f(m))^2$$ What about following?

Assume strictly monotone increasing function; such that $f:N^{+}\to N^{+}$, $h$ for all $n\in N^{+}$, $$f(f(f(n)))=f(f(n))\cdot f(n)\cdot n^{2015}$$

Prove or disprove:$f(n)=n^{13}$

Put $n=1,f(1)=m$ $$f(f(m))=mf(m)$$ Put $n=m$, $$f(f(f(m)))=f(f(m))f(m)m^{2015}\Longrightarrow f(mf(m))=m^{2016}(f(m))^2$$ What about following?

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user225250
user225250

Assume strictly monotone increasing function; such that $f:N^{+}\to N^{+}$, $h$ for all $n\in N^{+}$, $$f(f(f(n)))=f(f(n))\cdot f(n)\cdot n^{2015}$$

Assume strictly monotone increasing function; such that $f:N^{+}\to N^{+}$, $h$ for all $n\in N^{+}$, $$f(f(f(n)))=f(f(n))\cdot f(n)\cdot n^{2015}$$

Prove or disprove:$f(n)=n^{13}$

Prove or disprove:$f(n)=n^{13}$

Put $n=1,f(1)=m$ $$f(f(m))=mf(m)$$ Put $n=m$, $$f(f(f(m)))=f(f(m))f(m)m^{2015}\Longrightarrow f(mf(m))=m^{2016}(f(m))^2$$ What about following?

Assume strictly monotone increasing function; such that $f:N^{+}\to N^{+}$, $h$ for all $n\in N^{+}$, $$f(f(f(n)))=f(f(n))\cdot f(n)\cdot n^{2015}$$

Prove or disprove:$f(n)=n^{13}$

Put $n=1,f(1)=m$ $$f(f(m))=mf(m)$$ Put $n=m$, $$f(f(f(m)))=f(f(m))f(m)m^{2015}\Longrightarrow f(mf(m))=m^{2016}(f(m))^2$$ What about following?

Assume strictly monotone increasing function; such that $f:N^{+}\to N^{+}$, $h$ for all $n\in N^{+}$, $$f(f(f(n)))=f(f(n))\cdot f(n)\cdot n^{2015}$$

Prove or disprove:$f(n)=n^{13}$

Put $n=1,f(1)=m$ $$f(f(m))=mf(m)$$ Put $n=m$, $$f(f(f(m)))=f(f(m))f(m)m^{2015}\Longrightarrow f(mf(m))=m^{2016}(f(m))^2$$ What about following?

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N. F. Taussig
N. F. Taussig
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Jyrki Lahtonen
Jyrki Lahtonen
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user225250
user225250