F.E. (Z to Z)

For discussing Olympiad Level Algebra (and Inequality) problems
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Phlembac Adib Hasan
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F.E. (Z to Z)

Unread post by Phlembac Adib Hasan » Tue Nov 12, 2013 9:06 pm

Determine all functions $f : \mathbb{Z} \to \mathbb{Z}$ such that \[ f(a+b)^3 -f(a)^3 -f(b)^3 = 3f(a)f(b)f(a+b) ,\] for all $a,b \in \mathbb{Z}.$
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Tahmid Hasan
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Re: F.E. (Z to Z)

Unread post by Tahmid Hasan » Thu Nov 14, 2013 2:19 pm

First we look for constant solutions: $c^3-c^3-c^3=3c.c.c \Rightarrow c=0$, which is indeed a solution.
Let $P(a,b) \Rightarrow f(a+b)^3-f(a)^3-f(b)^3=3f(a)f(b)f(a+b)$.
$P(0,0) \Rightarrow f(0)=0$
$P(a,-a) \Rightarrow f(-a)=-f(a)$.
$P(a+1,-a) \Rightarrow f(a+1)^3=f(a)^3+3f(a+1)f(a)f(1)+f(1)^3$. Let's denote this $Q(a)$.
Let $f(a+1)=x,f(a)=y,f(1)=z$.
So $Q(a) \Rightarrow x^3=y^3+z^3+3xyz \Rightarrow (x-y-z)(x^2+y^2+z^2-xy+yz-zx)=0$
So $x=y+z,\frac{y+z \pm \sqrt{-2(y+z)^2-(y-z)^2}}{2}$.
The latter is an integer only when $y=z=0$.
If $f(1)=0$, then $Q(a) \Rightarrow f(a+1)=f(a) \forall a \in \mathbb Z$. So by induction $f(a)=0 \forall a \in \mathbb Z$.
Now we assume $f(1) \neq 0$.
Then $f(a+1)=f(a)+f(1)$, by induction we have $f(a)=af(1) \forall a \in \mathbb N$.
And since $f(-a)=-f(a)$, we conclude $f(a)=af(1) \forall a \in \mathbb Z$.
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Re: F.E. (Z to Z)

Unread post by Phlembac Adib Hasan » Thu Nov 14, 2013 4:33 pm

Tahmid Hasan wrote:So $Q(a) \Rightarrow x^3=y^3+z^3+3xyz \Rightarrow (x-y-z)(x^2+y^2+z^2-xy+yz-zx)=0$
এই ইকুয়েশনের সলিউশনে কোথাও ভুল হয়েছে। কারণ ৩ মডের আরেকটা সলু পাওয়া যায়।
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Re: F.E. (Z to Z)

Unread post by Tahmid Hasan » Thu Nov 14, 2013 6:02 pm

আমিও এইরকম এটা সলুশন সেট পাইছিলাম কিন্তু এই সলুশনের বাগটা খুঁজে পাইতেছি না। কাল টেস্ট শুরু, তাই একটু সময় সংকটে আছি, এজন্যে খালি স্কেচটা দিচছি। একটু আলাদা অ্যাপ্রোচঃ
$Q(a) \Rightarrow f(a+1)^3=f(a)^3+3f(a+1)f(a)f(1)+f(1)^3$.
$Q(1) \Rightarrow f(2)=2f(1),-f(1)$.
C1: $f(2)=2f(1)$.
$Q(2) \Rightarrow f(3)=3f(1)$.
Now use induction on $Q(a)$ to prove $f(a)=af(1)$.
[You'll get something like $(f(a+1)-f(a)-f(1))(f(a+1)^2+(f(a)+f(1))f(a+1)+f(a)^2-f(a)+f(1))=0$ which will provide the desired solution].
C2.$f(2)=-f(1)$.
$Q(2) \Rightarrow f(3)=0$.
$Q(3) \Rightarrow f(4)=f(1)$.
$Q(4) \Rightarrow f(5)=2f(1),-f(1)$.
$P(3,2)$ will provide a contradiction for $f(5)=2f(1)$.
Now claim $f(3k)=0,f(3k+1)=f(1),f(3k+2)=-f(1), k \in \mathbb{N_0}$.
Use induction to prove this.[Prove this using the ideas we used to prove the base cases].
Then use $f(-a)=-f(a)$ to expand it to $\mathbb Z$.
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Re: F.E. (Z to Z)

Unread post by mutasimmim » Sat Sep 06, 2014 10:26 am

Pardon me if it's a silly question but what does $f(a+b)^3$ mean? Is it $f[(a+b)^3]$ or $[f(a+b)]^3$?

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Re: F.E. (Z to Z)

Unread post by Phlembac Adib Hasan » Sat Sep 06, 2014 9:27 pm

Obviously, the later. To me, writing $f[(a+b)^3]$ as $f(a+b)^3$, is a crime ftw :)
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