Recent content by al-mahed

hi math-grl so what you want is to find the n's such that \varphi(n_1)=m_1 \varphi(n_2)=m_2 \varphi(n_3)=m_3 \varphi(n_4)=m_4 ... knowing only the m's, correct? there is a conjecture related to it, although what you want is far more difficult than the conjecture...

what you mean by "inverse"? do you mean multiplicative inverse ab=1 such that b = a^-1? I think you want to know if a given number is a phi of one or more numbers, and I think there is no general way to do it yet, only by hand for instance, given 14, is it a phi of at least one number? no...

Popitar, I'm not a teacher, I don't even have a major in mathematics, I'm just a curious guy, so you have to check all this stuff with your teacher. Anyway, I think you are having trouble to understand the basic concepts of proof, but don't worry, you'll get it soon enough (it is quite...

write x-1=a, then k=ma=m(x-1) for some integer m, since x-1| k, it yields x=a+1 x^[k-1]+x^[k-2]+...+x+1 = (a+1)^[k-1]+(a+1)^[k-2]+...+(a+1)+1 = a*S + k=(x-1)S+m(x-1), since you'll have k 1's, and the rest of it multiplied by a

correct, k elements, now you must show some work on it, grab a coffe (if you like) and think about it hint: what means "x-1 divide k" in terms of euclidian form a=qb+r? how could you WRITE it down as an equation? and how to use it? you already saw by yourself you need to prove that x-1 divide...

how many elements you have in x^[k-1]+x^[k-2]+...+x+1 ?

oops, in fact it is (...)and by Lagrange we know that either k=(n-1)\cdot \varphi{(n-1)} or k=(n-1)\cdot \varphi{(n-1)}w popitar, try to rewrite and "play" with the equations you know related to it, factoring x^k-1 is a way, did your teacher give any similar problem with solutions?

this is an implication "if and only if", <==> I'll do the part <==, then you try to do the part ==> prove that if n-1|k, then (n-1)^2|n^k-1: proof: first what is \varphi{([n-1]^2)} ?? it is (n-1)^{2-1}\cdot \varphi{(n-1)}=(n-1)\cdot \varphi{(n-1)} now, by euler, as gcd(n-1,\ n)=1 then...

you may want to check \varphi{([n-1]^2)}

yes, is it the only one?

find solutions where a) p is a prime; b) p is composite; \varphi{(p)}- \varphi{(p-1)}= [\varphi{(\varphi{(p)})}]^3 ps: it is not homework

don't worry anyway, I was thinking on a similar problem, to prove (or disprove) the statement:p^5 never divides a^p+b^p, that is, the contrary proplem (a superior limit), ohh, and gcd(a,p)=(b,p)=1

yes, that's exactly what I told to Robert, see the link I pasted for him above anyway ipetrich already closed the case quite elengantly at page 1

just to add that I put this question in a math onlympiad forum and someone gave a much faster, easier and better answer, though the first part complete my proof a^6\equiv\ 1\ mod\ 7 \ ==>\ (a^6)^{15}=a^{90}\equiv\ 1\ mod\ 7 a^{10}\equiv\ 1\ mod\ 11 \ ==>\ (a^{10})^{9}=a^{90}\equiv\ 1\ mod\...

I must confess I'm still studying abstract algebra, but I think I understand what you're saying, I'm assuming the problem initial hipothesis is true, and it is true since yields euler's totient function at the end of the day (for n>1), and it is true for n=1 by the verification via PARI aid...