Search found 79 matches
- Sat Feb 14, 2015 6:31 am
- Forum: Number Theory
- Topic: Iran TST 2011 D4 P3
- Replies: 1
- Views: 2540
Re: Iran TST 2011 D4 P3
My Solution: For any odd prime $p$, we give the legendre symbol $(\dfrac{1/2}{p})$ the usual values $1$ and $-1$, respectively whether there exists an integer $x$ with $2x^2 \equiv 1$ (mod $p$) or do not exist such $x$. We define $T = ${$p \in P-${$2$}$: (\dfrac{1/2}{p}) = -1$}. Here $P$ is the set ...
- Tue Feb 03, 2015 8:22 pm
- Forum: Number Theory
- Topic: Cool NT
- Replies: 1
- Views: 2445
Re: Cool NT
My solution: Call a nonempty set S of integers ‘good’ if $\forall x,y \in S \Rightarrow 1-(x+y) \in S$. Lemma: If S is good and $\exists p,q \in S: p-q = 3$, then $S = \mathbb{Z}$. Proof: Let arbitrary $x,y,z \in S$. Then $1-(x+y) \in S \Rightarrow 1-((1-(x+y))+z) = (x+y) – z \in S…(1)$. Also $(1-2x...
- Mon Feb 02, 2015 10:00 pm
- Forum: Number Theory
- Topic: Romania TST 2013
- Replies: 1
- Views: 2508
Re: Romania TST 2013
My solution: Claim: For each positive integer n, there exists an infinite sequence of nonempty finite sets $A_1,A_2,….$ of positive integers with $n < min(A_1)$ and for all i; $Max(A_i) < min(A_{i+1})$ , and $\sum_{s \in A_i}\dfrac{1}{s} = \dfrac {1}{n}$. Proof: In fact, it is well-known that for an...
- Thu Jan 29, 2015 8:11 pm
- Forum: Higher Secondary Level
- Topic: Infinity*zero= ??
- Replies: 5
- Views: 5182
Re: Infinity*zero= ??
Actually ''en.wikipedia.org/wiki/Cardinal_number'' may help you understand arithmetic's of infinity. Generally you can 'classify' sets to be 'equivalent' if they posseses a bijection. In this notion; 'infinity' is not a single element; there are variations of infinity[for example, natural number's s...
- Thu Jan 01, 2015 11:47 pm
- Forum: Number Theory
- Topic: a diophantine analogy to square's(self-made)
- Replies: 0
- Views: 1901
a diophantine analogy to square's(self-made)
Determine all pair of coprime positive integers $a,b$ so that for all positive integers $n$;
$a^n+b^{n+1}$ is a perfect square.
$a^n+b^{n+1}$ is a perfect square.
- Thu Dec 25, 2014 1:09 am
- Forum: International Mathematical Olympiad (IMO)
- Topic: IMO Shortlist 2004 N6
- Replies: 1
- Views: 2847
Re: IMO Shortlist 2004 N6
$P_1 = 1, P_2 = 1$. Let $n > 2$. Set $S_n = (x \in Z_n : x^2 = [1])$. Let $a(n) = |S_n|$. The idea is to note that $x \in S_n$ and $n-x \in S_n$ are equivalent. Note that all classes in $S_n$ are coprime with $n > 2$. So in case $n$ is even, $\frac{n}{2}$ is not in $S_n$. So of course, $a(n)$ is eve...
- Wed Dec 24, 2014 11:52 pm
- Forum: Number Theory
- Topic: Infinite Arithmetic Progression
- Replies: 1
- Views: 2141
Re: Infinite Arithmetic Progression
I think all n works. Obviously n = 1 works. Suppose n > 1. Now fix any prime p and set $s_m = p^{n-1}(p^{n-1}m + 1)$. Obviously $S = (s_m, m \in \mathbb{N})$ is an arithmetic progression with initial term $s_1$ and difference $s_2-s_1$. As $\tau$ is a multiplicative function, it follows that $S$ sat...
- Thu Dec 18, 2014 11:17 pm
- Forum: Algebra
- Topic: Existence of Multiplicative Inverse in a Commutative Ring
- Replies: 2
- Views: 2728
Re: Existence of Multiplicative Inverse in a Commutative Rin
In the polynomial ring $\mathbb{Q}[x]$ of field $\mathbb{Q}$, consider the principal ideal $I = <P(x)>$ generated by P[P is the minimal polynomial for $\alpha$]. Also consider the quotient ring $M = \mathbb{Q}[x]/I$; which is in fact a field since $P$ is irreducible. Now note that the map $H : M \ri...
- Wed Dec 17, 2014 8:48 pm
- Forum: Number Theory
- Topic: Can you find what to invoke?
- Replies: 2
- Views: 2668
Re: Can you find what to invoke?
Any number of the form $4x+3$ works. This is because: $n^{\tau(n)}$ is a perfect square for all positive integers n.[If n is itself perfect square, done. Else $\tau(n)$ is even, and done.] And $x^2+y^2$ is 0 or 1 or 2 mod 4.
Note: I am not feeling good....
Note: I am not feeling good....
- Sun Nov 23, 2014 4:37 pm
- Forum: Number Theory
- Topic: function for quadratic residue
- Replies: 2
- Views: 3072
function for quadratic residue
Determine all integers(ALL) $a $ such that there exists an integral valued function $f $ depending on $a$, defined for all sufficiently large primes $ p$ (domain depends on $a$) ; $f(p)^2-a$ is divisible by $ p$.