Search found 79 matches
- Fri Nov 07, 2014 2:44 pm
- Forum: Number Theory
- Topic: NT from Vietnam 2005
- Replies: 3
- Views: 3377
Re: NT from Vietnam 2005
$n = 0$ is impossible. So let for advantage, $x,y,n > 0$. Assume $y \geq x$. Now $x!(1+ \frac{y!}{x!}) = n!3^n…..(1) $ First case: First let $x \geq n$. Then $d = \frac{x!}{n!} (1+ \frac{y!}{x!}) = 3^n$ is odd. So $n+1 \geq x$ and $y > x$. Subcase 1: If $x = n+1; (n+1)(1+ \frac{y!}{(n+1)!}) = 3^n$. ...
- Mon Nov 03, 2014 10:56 pm
- Forum: Number Theory
- Topic: POLISH MATHEMATICAL OLYMPIAD
- Replies: 1
- Views: 2411
Re: POLISH MATHEMATICAL OLYMPIAD
We consider integers $p,q,r\geqslant 2$ and define $f(p,q) = pq-(p+q) \in \mathbb{Z}$. $f$ is symmetric on $p,q$. Note that $f(p+1,q)-f(p,q) = q-1 \geqslant 1…(i)$. As $f(2,2) = 0$; the following facts are easy with $(i)$: 1. $pq \geqslant p + q$; equality iff $p = q = 2$. 2. $pq \geqslant p + q + 1...
- Sun Nov 02, 2014 6:14 pm
- Forum: Number Theory
- Topic: Divisibility 3
- Replies: 1
- Views: 2442
Re: Divisibility 3
We are done if $xy = 0$. So let $x,y$ be positive integers. Rewrite as $x^2 – (n^2 – 1)y^2 = 1….(i)$, where $d = (n^2 – 1) $ is a nonsquare positive integer. We now know from the theory of pell’s equations that (i) has a nontrivial solution (x,y) in positive integers. Let $x_0 + y_0\sqrt{d}$ be the ...
- Sun Nov 02, 2014 5:20 pm
- Forum: Number Theory
- Topic: Divisors...
- Replies: 1
- Views: 2381
Re: Divisors...
Write $m = 2^{s} \prod^{i = k}_{i=1} {p_{i}}^{e_{i}}$ be the prime factorization of $m; s, k$ are nonnegative; all $p_i$’s divide m. From the given relation: $(3s + 2) \prod (3e_i+1) = 2^{s+1} \prod {p_i}^{e_i} = 2m….(1)$ It tells that $3$ does not divide $m$. Now simply note that 1. $2^x \geqslant ...
- Sat Oct 25, 2014 1:50 pm
- Forum: Algebra
- Topic: China 2011 tst quiz-1 problem 1
- Replies: 1
- Views: 2697
Re: China 2011 tst quiz-1 problem 1
The given equation is denoted $(M)$. Let $f(0) = c$. $(M) f(y),y \rightarrow f(y^n + f(y)) = \frac {c}{2} ….(i)$ $(i), (M) \rightarrow f(x – f(y)) = f(x + y^n) + \frac{c}{2} ….(M_1)$ $(M_1) x + f(a), a(\forall a \in \mathbb{R})$ $\rightarrow f(x + (f(a) + a^n)) = f(x) - \frac{c}{2}…(1)$ $(M_1) x – b...
- Thu Oct 23, 2014 9:37 pm
- Forum: Algebra
- Topic: Beautiful FE
- Replies: 1
- Views: 2787
Re: Beautiful FE
The given equation is denoted $(M)$. First we note that if $f(a) = f(b)$, $(M)$ implies $nb = f(a+b+f(b)) – f(a) = f(b+a+f(a))- f(b) = na \Rightarrow a = b$, as $n \neq 0$. $\rightarrow f$ is injective. Now $(M) y = 0 \rightarrow f(x+f(0)) = f(x) \rightarrow f(0) = 0$. ($M) x=0 \rightarrow f(y+f(y))...
- Thu Oct 23, 2014 5:45 pm
- Forum: Number Theory
- Topic: Inequality with Prime factorisation(TST problem)
- Replies: 0
- Views: 1913
Inequality with Prime factorisation(TST problem)
Let $p$ be a sufficiently large prime. Suppose $(p-1)^p$ has prime factorisation $\prod\limits^n_{i=1} p_i^{e_i}$.
Prove that $\sum p_ie_i \geqslant \frac{p^2}{2}$
Prove that $\sum p_ie_i \geqslant \frac{p^2}{2}$
- Mon Oct 20, 2014 7:37 pm
- Forum: International Mathematical Olympiad (IMO)
- Topic: IMO Marathon
- Replies: 184
- Views: 115554
Re: IMO Marathon
My solution may contain typing mistakes. Anyway, nice problem. Solution: In this solution, I assume some lines are concurrent iff either all are parallel or all have a finite intersection point. $AA’, BB’, CC’$ are concurrent, say at $X$; as they are respectively radical axises of the pairs $\omega_...
- Sun Oct 19, 2014 10:47 pm
- Forum: Algebra
- Topic: Turkey TST 2014
- Replies: 2
- Views: 3556
Re: Turkey TST 2014
Main equation is denoted $(M)$. $f(1) = k$. $(M)x = 0 \rightarrow f(f(y)+1) = k^2+y…(ii)$. So f is bijective. $(ii) y-k^2 \rightarrow f(f(y-k^2)+1) = y….(ii-)$ $(M)(x,y)$ vs. $(-x,y) \rightarrow f(x+1)^2 – f(1-x)^2 = 4x ….(i)$ $(ii-) y = k^2+1 \rightarrow f(k+1) = k^2+1….(vii)$ $(ii)y = k+1 \rightar...
- Wed Oct 01, 2014 2:41 pm
- Forum: Number Theory
- Topic: Quadratic Residues Modulo Prime
- Replies: 6
- Views: 4404
Re: Quadratic Residues Modulo Prime
Another solution: Consider the two sets $X,Y$ of residue classes (mod p) as defined next: $X$ contains squares of classes through $0$ to $\frac{p-1}{2}$. $Y$ contains classes of form $(r-t)$; for t$\in X$. It is obvious that there are exactly $\frac{p+1}{2}$ elements in both $X$ and $Y$. As $A \cup ...