yey, my problem strikes again :P this may be a little unfair, as this has been posted on the forum also (I know that I posted it , but also can be found in the RMC booklet last of last year). check out this topic http://www.mathlinks.ro/viewtopic.php?t=3886&highlight=rom%2A+tst+2003

A nice problem has to be shared as often as possible As I said in another topic, I didn't take part of the selection of the problems. Otherwise I would eliminate it because of mathlinks and that a similar one (without the restriction that $A \neq M$ ) has been proposed and solved (by me!!!) on the forum of the french candidates.....When I saw this problem on the exam paper monday (it was for the thusday) I told our team leader Claude Deschamps about it (he turns into green) but it was too late because the exams is not centered at one place but is on different towns in France and we have to use people who are absolutely not involved into olympiads.... Fortunately, none of the candidates seems to have remind it. Pierre.

my initial problem was without the restriction $A\neq M$, but I decided to make it a little harder for the TST

I think a very interesting problem arises if instead of the product -1 we consider the product+1. It seems to be true too... but is probably harder. psykyk

well actually, that question was the first question. It derived from Euclid's proof of the existance of an infinity of primes. indeed it is harder.

hi guys i m new member , i m from morocco , i already participate in IMO 01 and IMO 02 ,this is what i did about the problem ( just excuse my mathematical english ) i think it's evident to say that M is infinite! so bye induction if p1 , p2 ,...,pn are the n first prime numbers and if they belongs to M. let p be the next prime as we know M is infinite so there is at least a class of equivalence modulo p which contains an infinity of primes from M, let's take p-1 prime from this class ,this subset of M verify (by fermat's theoreme n^(p-1) congru 1 (mod p))the condition so p is in M. and so M eats every prime sorry i know u did th problem but i just wanna participate by the way valentin i remember u [/tex][/i]

You are welcome In another hand, I think that proving the set is infinite is the hardest part of the problem (you would loose at least 4 points for that , Claude was pityless). Pierre.

pbornsztein wrote: You are welcome In another hand, I think that proving the set is infinite is the hardest part of the problem (you would loose at least 4 points for that , Claude was pityless). Pierre. I was too at the TST marking in Romania ... :D

i will check my papers guys sorry! it seems that i forgot the A <> M

lets p_1,....,p_n be the primes in M, with p_1<...<p_n A={p_1,p_2,....,p_n-1} the product X(A) of the elements of A : X(A)-1 is divisibl byp_n so let r be the exposant X(A)=1+{p_n}^{r} or X(A)=(1+p_1^{x}p_n^{y})p_1 y couldn't be 1 if not p_1:equiv: 1(mod p_n) wich is absurde so X(A) :equiv: 1(mod p_1) now A={p_2,p_3,....,p_n} p_n :equiv: 1(mod p_1) so p_n :equiv: 1(modX(A)) absurde so the M is finite [/tex]

sorry frac{X(A)}{p_1} :equiv: 1(mod p_1) sorry !!