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Re: $a\equiv b$ whenever $F(a)\equiv F(b) (mod p)$

oh, I missed it actually. My bad, sorry.
by SANZEED
Thu Oct 02, 2014 1:33 pm
 
Forum: Number Theory
Topic: $a\equiv b$ whenever $F(a)\equiv F(b) (mod p)$
Replies: 8
Views: 1186

Re: $a\equiv b$ whenever $F(a)\equiv F(b) (mod p)$

For \(p=2\) we have \(F(n)=1\) for all \(n\) so I guess the question asks to prove for odd primes. Notice that \[F(n) = \sum_{k=1}^{p-1} kn^{k-1} = \sum_{k=1}^{p-1} \dfrac{\text{d}}{\text{d}n} n^k=\dfrac{\text d}{\text d n}\sum_{k=1}^{p-1} n^k=\dfrac{\text d}{\text d n}\left(\dfrac{n^p-n}{n-1}\righ...
by SANZEED
Thu Oct 02, 2014 1:02 pm
 
Forum: Number Theory
Topic: $a\equiv b$ whenever $F(a)\equiv F(b) (mod p)$
Replies: 8
Views: 1186

Re: 2-adic valuation of $n!$

Let $j(n,p)$ denote the highest power of $p$ that divides $n!$. Then, it is a well-known fact that, $j(n,2)=\displaystyle\sum_{k=1}^{\infty}\lfloor \frac{n}{2^{k}}\rfloor=\lfloor \frac{n}{2}\rfloor+\lfloor \frac{n}{4}\rfloor+\lfloor \frac{n}{8}\rfloor+...$. Now, it is also well-known that $1=\dfrac{...
by SANZEED
Wed Oct 01, 2014 12:11 pm
 
Forum: Number Theory
Topic: 2-adic valuation of $n!$
Replies: 2
Views: 533

Re: Binoial congruence(mod p)

I think there is a typo, because your statement is false for $p=3$. May be the correct statement is $2p\choose p$ $\equiv 2(mod p^{2})\forall p=$ prime. I have a proof for this. Let us choose a set of $p$ boys and girls from a set of $p$ boys and $p$ girls. Firstly, this can be done in $2p\choose p$...
by SANZEED
Mon Sep 29, 2014 5:49 pm
 
Forum: Number Theory
Topic: Binoial congruence(mod p)
Replies: 2
Views: 602

Re: Functional Equation( Japan final round 2008)

Of course $f(x)=0$ is the unique constant solution. We assume $f$ isn't constant and denote by $P(X,Y)$ the assertion that $f(x+y)f(f(x)-y)=xf(x)-yf(y)$. $P(x,0)$ implies $f(x)f((f(x))=xf(x)$ and so $f(f(x))=x$. $P(0,y)$ implies $f(y)f(f(0)-y)=-yf(y)$ and so $f(f(0)-y)=-y$, so $f(f(f(0)-y))=f(-y)$ ...
by SANZEED
Mon Sep 29, 2014 1:00 am
 
Forum: Algebra
Topic: Functional Equation( Japan final round 2008)
Replies: 6
Views: 887

Re: Functional Equation( Japan final round 2008)

@Turzo, sorry, actually I didn't have time to post this case, and because it's a bit obvious, I left it while posting. Anyway, here is the third case: $\textbf{Case 3:} \exists a,b$ such that $f(a+f(a))=0$ and $f(b+f(b))$ is non-zero, where both $a,b$ are non-zero reals. According to our previous re...
by SANZEED
Mon Sep 29, 2014 12:55 am
 
Forum: Algebra
Topic: Functional Equation( Japan final round 2008)
Replies: 6
Views: 887

Re: Functional Equation( Japan final round 2008)

Let us denote the given statement by $P(x,y)$. Then, $P(x,f(x))\Rightarrow f(x+f(x))f(0)=xf(x)-f(x)f(f(x))$. $P(x,0)\Rightarrow f(x)f(f(x))=xf(x)$. So, $f(x+f(x))f(0)=0 \forall x\in \mathbb{R}$. So we have two cases here. $\textbf{Case 1:}$ When $f(x+f(x))=0 \forall x\in \mathbb{R}$. $P(x+f(x),-x)\R...
by SANZEED
Sun Sep 28, 2014 4:41 pm
 
Forum: Algebra
Topic: Functional Equation( Japan final round 2008)
Replies: 6
Views: 887

Re: IMO Marathon

My solution to 37: Let $P\in AC$ such that $AP=AB$. Let $BP\cap AI=X_0$. Let $EX_0$ meet $\odot DEF$ at $D_0\neq E$. Clearly $D_0$ is on the other side of $BP$ than $F$. Now, $AI$ is the perpendicular bisector of $EF$ and $BP$. Again, $\angle FD_0X_0=\angle ED_0F=\angle AFE=\angle ABP=\angle FBX_0$...
by SANZEED
Sat Sep 27, 2014 6:49 pm
 
Forum: International Mathematical Olympiad (IMO)
Topic: IMO Marathon
Replies: 184
Views: 23432

Re: IMO Marathon

@Jakaria, $f(x)=-x^{2}$ is not a valid solution. Check it please. My solution: I am skipping obvious calculations here. Let us denote the given statement by $P(x,y)$. Then, $P(x,0)\Rightarrow f(f(x))=f(x^{2})-2f(0)$ $P(1,1)\Rightarrow f(f(1)-1)=f(1)+f(1)-2f(1)=0$. Let $b=f(1)-1$. Then $f(b)=0$. So, ...
by SANZEED
Fri Sep 26, 2014 10:33 pm
 
Forum: International Mathematical Olympiad (IMO)
Topic: IMO Marathon
Replies: 184
Views: 23432

Re: IMO Marathon

I think this marathon should be revived. Also, post a solution to problem 35, Asif, if you have it. :|

Problem 36
Find all functions $f:\mathbb{R}\rightarrow \mathbb{R}$ such that $\forall x,y\in \mathbb{R}$,
$f(f(x)-y^{2})=f(x^{2})+y^{2}f(y)-2f(xy)$.
by SANZEED
Thu Sep 25, 2014 11:38 pm
 
Forum: International Mathematical Olympiad (IMO)
Topic: IMO Marathon
Replies: 184
Views: 23432
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