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Dynic

#1 h Yesterday at 5:10 PM

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Find all function $f:\mathbb{Z}\to\mathbb{Z}$ satisfy all conditions below:
i) $f(n+1)>f(n)$ for all $n\in \mathbb{Z}$
ii) $f(-n)=-f(n)$ for all $n\in \mathbb{Z}$
iii) $f(a^3+b^3+c^3+d^3)=f^3(a)+f^3(b)+f^3(c)+f^3(d)$ for all $n\in \mathbb{Z}$

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#2 h Yesterday at 6:15 PM

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Dynic wrote:

Find all function $f:\mathbb{Z}\to\mathbb{Z}$ satisfy all conditions below:
i) $f(n+1)>f(n)$ for all $n\in \mathbb{Z}$
ii) $f(-n)=-f(n)$ for all $n\in \mathbb{Z}$
iii) $f(a^3+b^3+c^3+d^3)=f^3(a)+f^3(b)+f^3(c)+f^3(d)$ for all $n\in \mathbb{Z}$

Trivial by the sum of four cubes conjecture.

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#3 h Yesterday at 6:30 PM

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jasperE3 wrote:

Dynic wrote:

Find all function $f:\mathbb{Z}\to\mathbb{Z}$ satisfy all conditions below:
i) $f(n+1)>f(n)$ for all $n\in \mathbb{Z}$
ii) $f(-n)=-f(n)$ for all $n\in \mathbb{Z}$
iii) $f(a^3+b^3+c^3+d^3)=f^3(a)+f^3(b)+f^3(c)+f^3(d)$ for all $n\in \mathbb{Z}$

Trivial by the sum of four cubes conjecture.

I think this conjecture hasn't been proven yet.

This post has been edited 1 time. Last edited by CHESSR1DER, Yesterday at 6:46 PM

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#4 h Yesterday at 7:48 PM

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Dynic wrote:

Find all function $f:\mathbb{Z}\to\mathbb{Z}$ satisfy all conditions below:
i) $f(n+1)>f(n)$ for all $n\in \mathbb{Z}$
ii) $f(-n)=-f(n)$ for all $n\in \mathbb{Z}$
iii) $f(a^3+b^3+c^3+d^3)=f^3(a)+f^3(b)+f^3(c)+f^3(d)$ for all $n\in \mathbb{Z}$

The second condition is not necessary, because it follows from iii): if we take $a=b=c=d=0,$ we get $f(0)=4f^3(0),$ so $f(0)=0.$ Then, by taking $b=-a$ and $c=d=0$ we get $f^3(-a)=-f^3(a),$ and so $f(-a)=-f(a).$

By taking $(a,b,c,d)=(1,0,0,0)$ in iii), we get $f(1)=f^3(1),$ which combined with $f(1)>f(0)=0$ leads to $f(1)=1.$

Taking $(a,b,c,d)=(1,1,1,1)$ leads to $f(4)=4.$ Then taking $(a,b,c,d)=(4,4,4,4)$ leads to $f(4^4)=4^4.$

By repeating this process, we get an unbounded sequence $(x_k)$ for which $f(x_k)=x_k.$ Combining this with the fact that $f$ is odd and strictly increasing, we obtain $f(n)=n$ for every $n.$

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Dynic

#5 h Today at 12:48 AM

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Filipjack wrote:

Dynic wrote:

Find all function $f:\mathbb{Z}\to\mathbb{Z}$ satisfy all conditions below:
i) $f(n+1)>f(n)$ for all $n\in \mathbb{Z}$
ii) $f(-n)=-f(n)$ for all $n\in \mathbb{Z}$
iii) $f(a^3+b^3+c^3+d^3)=f^3(a)+f^3(b)+f^3(c)+f^3(d)$ for all $n\in \mathbb{Z}$

The second condition is not necessary, because it follows from iii): if we take $a=b=c=d=0,$ we get $f(0)=4f^3(0),$ so $f(0)=0.$ Then, by taking $b=-a$ and $c=d=0$ we get $f^3(-a)=-f^3(a),$ and so $f(-a)=-f(a).$

By taking $(a,b,c,d)=(1,0,0,0)$ in iii), we get $f(1)=f^3(1),$ which combined with $f(1)>f(0)=0$ leads to $f(1)=1.$

Taking $(a,b,c,d)=(1,1,1,1)$ leads to $f(4)=4.$ Then taking $(a,b,c,d)=(4,4,4,4)$ leads to $f(4^4)=4^4.$

By repeating this process, we get an unbounded sequence $(x_k)$ for which $f(x_k)=x_k.$ Combining this with the fact that $f$ is odd and strictly increasing, we obtain $f(n)=n$ for every $n.$

Nice, thanks! We can also use this equality: $6k =(k+1)^3+(k-1)^3+(-k)^3+(-k)^3$

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ohiorizzler1434

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ohiorizzler1434

#6 h Today at 1:49 AM

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Great! It represents how solving a conjecture gives us many insightful algebraic identities!

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