## Search found 550 matches

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: 3461

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

oh, I missed it actually. My bad, sorry.
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: 3461

### 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...
Wed Oct 01, 2014 12:11 pm
Forum: Number Theory
Topic: 2-adic valuation of $n!$
Replies: 2
Views: 1513

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 $1=\dfrac{... Mon Sep 29, 2014 5:49 pm Forum: Number Theory Topic: Binoial congruence(mod p) Replies: 2 Views: 1623 ### Re: Binoial congruence(mod p) I think there is a typo, because your statement is false for$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$2p\choose p$... Mon Sep 29, 2014 1:00 am Forum: Algebra Topic: Functional Equation( Japan final round 2008) Replies: 6 Views: 2932 ### 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$f(f(f(0)-y))=f(-y)$... Mon Sep 29, 2014 12:55 am Forum: Algebra Topic: Functional Equation( Japan final round 2008) Replies: 6 Views: 2932 ### 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... Sun Sep 28, 2014 4:41 pm Forum: Algebra Topic: Functional Equation( Japan final round 2008) Replies: 6 Views: 2932 ### 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...
Sat Sep 27, 2014 6:49 pm
Topic: IMO Marathon
Replies: 184
Views: 63440

### 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, $\angle FD_0X_0=\angle ED_0F=\angle AFE=\angle ABP=\angle FBX_0$...
Fri Sep 26, 2014 10:33 pm
Topic: IMO Marathon
Replies: 184
Views: 63440

### 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, ...
Thu Sep 25, 2014 11:38 pm
I think this marathon should be revived. Also, post a solution to problem 35, Asif, if you have it. Find all functions $f:\mathbb{R}\rightarrow \mathbb{R}$ such that $\forall x,y\in \mathbb{R}$,
$f(f(x)-y^{2})=f(x^{2})+y^{2}f(y)-2f(xy)$.