20190101, 23:33  #23  
Sep 2017
3^{2}·11 Posts 
Quote:


20190102, 00:46  #24 
Feb 2017
Nowhere
7·641 Posts 
AFAIK there is no such thing as repetition of elements in a set. A set is not the same thing as an ordered tuple.
The two sets need not be disjoint, however. Clearly, adding a number to all the elements of one set, and subtracting it from all the elements of the other, creates two sets giving the same sums. So instead of [1,22] and [3,99] we could take [2,23] and [2,98]. Last fiddled with by Dr Sardonicus on 20190102 at 00:47 Reason: gixfin posty 
20190102, 00:58  #25  
"Rashid Naimi"
Oct 2015
Remote to Here/There
2^{2}×503 Posts 
Quote:
Thank you for the correction. 

20190102, 01:08  #26  
"Forget I exist"
Jul 2009
Dumbassville
2^{6}×131 Posts 
Quote:
Last fiddled with by science_man_88 on 20190102 at 01:13 

20190103, 18:58  #27 
"Kebbaj Reda"
May 2018
Casablanca, Morocco
83 Posts 
triplet with a pair
Until now, i only have a triplet with a pair :
A = {22, 97, 526}; B = {3, 99}; 22 + 3 = 25 Factor {{5^2}} 22 + 99 = 121 Factor {{11^2}} 97 + 3 = 100 Factor {{2^2},{5^2}} 97 + 99 = 196 Factor {{2^2},{7^2}} 526 + 3 = 529 Factor {{23^2}} 526 + 99 = 625 Factor {{5^4}} 
20190103, 19:31  #28 
"Kebbaj Reda"
May 2018
Casablanca, Morocco
83 Posts 
it is going well
it is going well: quadriplet with pair:
B = {3, 99, 4803, 45699} ; A = {97, 526}; 
20190104, 05:57  #29  
Romulan Interpreter
Jun 2011
Thailand
2·13·19^{2} Posts 
Quote:
25 > 1 100 > 1 etc. This problem has not much to program, is pure math, and quite simple actually. As said above, N must always be a perfect square. The sums can not be square free, because Alice would win in one shot, picking the whole number as the first divisor, and they can't be nonsquares, because then Alice will chose the first time in such a way to leave a perfect square, and she wins. For example, if \(N=2^a3^b5^c...\) then Alice picks first time the product of all primes which have the odd power, and then what is left is a perfect square. Now, the only left for you is to prove that if the number is perfect square, then Bob (the second picker) wins every time. There is a theorem which states that the only numbers with an odd number of divisors are the perfect squares (why? and why do you need an odd number of divisors?) anyhow this is not complicate to prove, but you don't need to go so deep, because there is a simple strategy to win: if N is perfect square, then any prime in N has its pair, and all Bob has to do is to pick the same product Alice picks, every time, leaving every time behind a perfect square. Start with a perfect square, end with a smaller perfect square, repeat. This ends in 1, and it is the "infinite descent" method, invented by Fermat, hehe  as members of mersenneforum, you should know that :razz: So, all the fuss is about finding two sets of numbers whose paired sums are always perfect squares. How difficult is that? (from the number of the persons who already solved the puzzle in just few hours, and from this current thread, you can see that the puzzle is trivial  so this is another skip this month). Last fiddled with by LaurV on 20190104 at 06:32 

20190104, 06:38  #30 
"Rashid Naimi"
Oct 2015
Remote to Here/There
2^{2}·503 Posts 
And how does any of that have to do with the text that he quoted from me and labeled as false?
Where is exactly the error in my post? Please make a direct reference to the false statement. Thanks. 
20190104, 07:17  #31  
Romulan Interpreter
Jun 2011
Thailand
2·13·19^{2} Posts 
Quote:
Oh? How come? I just said how Alice wins in this case... Last fiddled with by LaurV on 20190104 at 07:18 

20190104, 07:42  #32 
Jun 2003
11·449 Posts 
As I understand it, the sticking point for a1call is the "product of distinct primes" bit. According to him, a single prime is not a product  there should be at least two.
EDIT: Looks like they have updated "divides N by any divisor that is either a prime or a product of distinct prime numbers". That should take care of that. Last fiddled with by axn on 20190104 at 07:53 
20190104, 09:49  #33 
"Forget I exist"
Jul 2009
Dumbassville
2^{6}·131 Posts 

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