10 and it is possible for the following two sets. M={1,3,4,5,6,7,10,11,13,14} or {1,3,5,6,7,9,10,11,13,14}

Does somebody have a proof?

Assuming that praneeth's example is correct,we only need to establish that having eleven numbers in a such set is impossible. Consider two cases: 1.One is in the set:The following two numbers can not be in the set simultaneously: 1.$ 2$ and $ 8$ 2.$ 3$ and $ 12$ 3.$ 4$ and $ 9$ Also if $ 5$ and $ 10$ are both in the set it follows that neither of $ 2$ and $ 8$ can be in the set.The same claim holds for pair $ 7$ and $ 14$.It means that $ 2$ and $ 8$ are not in the set,while numbers: $ 1,5,6,7,10,11,13,14,15$ are in the set.Hence $ 12$ can not be in the set,but then $ 3\times 15\times 5=15^2$ 2.One is not in the set.I need to think on this case a little bit more

Well we know $ 1$ isn't in the set from the post above. If $ 2$ is in the set then our set cannot contain both elements of any of the following pairs: $ 3,6$ $ 4,8$ $ 5,10$ $ 6,12$ $ 7,14$ So if we have an 11-element set, $ 2$ cannot be in it. So $ 1$ and $ 2$ are not in the set, and our set cannot fully contain any of the following triples: $ 15,10,6$ $ 7,8,14$ $ 3,9,12$ So, that gives us 3 elements not in the set. With 1 and 2 not in the set either, we can have at most 10 elements in the set.

lol: It's pretty funny that our solution consists from three parts suggested by three different people...

orl wrote: $ M$ is a subset of $ \{1, 2, 3, \ldots, 15\}$ such that the product of any three distinct elements of $ M$ is not a square. Determine the maximum number of elements in $ M.$ Answer: 3. Example: $ M = \{1,4,9\}$. Assume $ |M| \geq 4$. Assume to the contrary that $ M$ contains at least 4 elements. Consider 4 elements $ a, b, c, d \in M$. Numbers $ abc$, $ abd$, and $ acd$ are perfect squares. Therefore, $ (abc)(abd)/(acd) = ab^2$ is a square of a rational number. Thus $ a$ is a perfect square. Similarly, $ b$, $ c$, and $ d$ are perfect squares. But there are only 3 distinct perfect squares between 1 and 15: 1, 4, and 9

Allnames, what happens when you add $ 5$ to your set?

Um, Allnames seems to be assuming that the problem involved sets such that the product of any three elements IS a perfect square, when it should be IS NOT a square. Anyway, I don't see what happens when you add $ 5$ to the set, except that the set will no longer satisfy either condition.

sorry,because of my bad English I tried it,and i found the result is $ 10$ But i think the problem that I solved is quite interesting

Hi! (excuse me if I make an english mistake , I'm portuguese) Isn't the right solution 11?? {1, 5, 6, 7, 9, 10, 11, 12, 13, 14, 15} See you on the next post

but $ 5,12,15$ notice tjhance solution

My mistake!

Any non bashy solution exists?

Then we form a subset of co-prime integers excluding squares,i.e. $$M=\{4,2,3,5,7,11,13\}$$We now then see the possibility of a square in the set too. $$M=\{1,2,3,5,7,9,11,13\}$$Then we search for square free integers. $$M=\{1,2,3,5,6,7,9,11,13\}$$Oof other than case bash?

I have decided to dedicate my first ever post on AoPS to this problem because after 12 long years, no one has found the solution $$M=\{3, 4, 5, 6, 7, 9, 10, 11, 13, 14\}$$And there is not 1, not 2, not 3, but $\textbf{4}$ missing solutions! $$M=\{1, 4, 5, 6, 7, 10, 11, 12, 13, 14\}$$$$M=\{1, 5, 6, 7, 9, 10, 11, 12, 13, 14\}$$$$M=\{4, 5, 6, 7, 9, 10, 11, 12, 13, 14\}$$Here is my solution. Let $S$ denote the set $\{1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 11, 12, 13, 14, 15\}$. We will first simplify $S$ to a multiset $S'$. Note that for each element of $S$, if we multiply or divide it by a perfect square, whether its product with $2$ other elements is a perfect square or not does not change. Thus, for each number $n$ in $S$, if there is a perfect square $k^2>1$ which divides it, we replace it with $\frac{n}{k^2}$. This yields $$S'=\{1, 2, 3, 1, 5, 6, 7, 2, 1, 10, 11, 3, 13, 14, 15\}$$or $$S'=\{1, 1, 1, 2, 2, 3, 3, 5, 7, 11, 13, 2\times 3, 2\times 5, 2\times 7, 3\times 5\}$$(I have prime factorised the numbers for easier reference) Define $M'$ similarly for $M$. We will prove that the upper bound of $|M'|$ is $10$ by proving we need to eliminate at least $5$ elements from $S'$ to satisfy the no perfect-square condition. Let $(x, y, z)\Rightarrow -1\hspace{0.2cm}(n)$ denote that because of the perfect-square triplet $(x, y, z)$, we have to eliminate at least $1$ element, leaving us with at most $n$ elements remaining. $(1, 1, 1)\Rightarrow -1\hspace{0.2cm}(14)$ $(2, 7, 2\times 7)\Rightarrow -1\hspace{0.2cm}(13)$ Now we diverge into $2$ cases. Case $1$: $2$ was removed because of $(2, 7, 2\times 7)$ Since there are $2$ $'2'$s, that means 2 elements were removed leaving us with $12$ elements. $(2\times 3, 2\times 5, 3\times 5)\Rightarrow -1\hspace{0.2cm}(11)$ $(1, 3, 3)\Rightarrow -1\hspace{0.2cm}(10)$ (We either have to remove both $'1'$s because of the double $3$, or remove one $'3'$.) Case $2$: $'2'$ was kept $(2, 5, 2\times 5)\Rightarrow -1\hspace{0.2cm}(12)$ $(2, 3, 2\times 3)\Rightarrow -1\hspace{0.2cm}(11)$ Note that for the triplet $(2, 3, 2\times 3)$, if $'3'$ was removed that means $2$ elements were removed instead of $1$, leaving us with $10$ elements and we are done. Otherwise, the element $2\times 3$ was removed. Then, $(1, 3, 3)\Rightarrow -1\hspace{0.2cm}(10)$ (Reason is similar to Case $1$) Thus $|M'|\le 10$ and we are done. As a bonus we will prove that $M'=\{1, 1, 3, 5, 7, 11, 13, 2\times 3, 2\times 5, 2\times 7\}$ is the only possible solution for $M'$. It then follows that there are $6$ solutions for $M$ - the $\{1, 1\}$ in $M'$ can become $\{1, 4\}$, $\{1, 9\}$, or $\{4, 9\}$ in $M$, and the $\{3\}$ can become $\{3\}$ or $\{12\}$. In Case $1$, if we have $|M'|=10$ it means we are keeping one $'3'$ and one $'5'$, which means out of the $3$ elements $2\times 3$, $2\times 5$, and $3\times 5$, we have to remove $3\times 5$. The resultant $M'$ is the solution above. In Case $2$, having $|M'|=10$ means that we must keep both $'2'$s and $2$ $'1'$s. But the triplet $(1, 2, 2)$ shows that this is impossible. We are done.

Dang this problem is bash... Here goes The answer is $\boxed{10}$. It is easy to see that this is achievable(and not too hard to find) that a set that works is 1,3,5,6,7,9,10,11,13,14. It is easy to check that this works by eliminating numbers based on common factor(for example, for $5$ to possibly be a perfect square, $10$ would also have to be part of the 3 element product, and then there a only a couple of cases to check). We now show that $10$ is optimal. Notice that at most $2$ elements in each of the sets {1,4,9}, {3,5,15}, {2,6,12}, and {7,8,14} can be used. This gives us a bound of $11$, still not strong enough. Now hound in on the two sets {2,6,12} and {3,5,15}. If we can somehow show that among these two sets, at most $3$ elements can be used, then we will be done. To show this, suppose we could choose at least $4$ elements among these two sets somehow. Then clearly this would be by choosing two from one set and two from another. The possible sets of four elements are {2,6,3,5} {2,6,3,15} {2,6,5,15} {2,12,3,5} {2,12,3,15} {2,12,5,15} {6,12,3,5} {6,12,3,15} {6,12,5,15}. Notice the first two sets are invalid because they contain {2,3,6}. Also notice that the 4th, 5th, 7th, and 8th cases are bad because they contain {3,12}, which means that none of elements in the set {1,4,9} can be used, already strong enough to prove the bound. Also notice that the 6th and 9th cases are bad because they contain {5,12,15} which product to a perfect square. Therefore, the only possible set that can have more than 10 elements is {2,6,5,15}. However, notice that this set means that in the set {7,8,14}, I must choose the 8, or else 7,14, and 2 would product to a perfect square, but with a 2 and an 8, then {1,4,9} would not work, already strong enough. So now we are done.

$\color{magenta} \boxed{\textbf{SOLUTION N1}}$ Let, $|M|$ denotes the number of elements in Set $M$ $\color{red} \textbf{CLAIM 1 :}$ If $|M|=11$ doesn't work then $|M|=12,13,14,15$ also doesn't work $\textbf{Proof :}$ Consider any $11$ elements from $M \ge 12$ We must find $3$ elements $a,b,c$ from $M$ such that $abc=x^2$ as we assumed $|M|=11$ doesn't work that is for every $|M|=11$ we will get three elements whose product is a perfect square $\blacksquare$ $\color{red} \textbf{CLAIM 2 :}$ $|M|=11$ doesn't work $\textbf{Proof :}$ Consider, $$A_1=(2,1,8)$$$$A_2=(2,3,6)$$$$A_3=(2,4,8)$$$$A_4=(2,5,10)$$$$A_5=(2,6,12)$$$$A_6=(2,7,14)$$$$A_7=(2,8,9)$$ Let, $2 \in M$ Then We have $6,8 \notin M$ Also one of $(5,10)$ and one of $(7,14)$ can't be in $M$ Then Consider $T={1,4,9}$ All of these must be in $M$ but $1\times 4\times 9=36=6^2$ Contradiction! So, $2 \notin M$ Now, Consider, $$B_1=(3,1,12)$$$$B_2=(3,2,6)$$$$B_3=(3,4,12)$$$$B_4=(3,5,15)$$$$B_5=(3,6,8)$$$$B_6=(3,9,12)$$ Let, $3\in M$ Then we have $6,12 \notin M$ And one of $(5,15)$ not in $M$ Again consider $T=(1,4,9)$ all of these must be in $M$ Contradiction! So, $3\notin M$ We want to find two more elements which can't be in $M$ Consider $$T=(1,4,9)$$$$N=(5,8,10)$$$$K=(5,12,15)$$$$L=(7,8,14)$$ Let, $5\in M \implies$ one of $(8,10)$ and one of $(12,15)$ can't be in $M$ But Then again consider $T=(1,4,9)$ Contradiction! So, $5 \notin M$ Now, Consider $T,L$ We we have $a\notin M, a\in T$ then $L=(7,8,14) \in M$ Contradiction! And if we have $b\notin M, b\in L$ then $T=(1,4,9) \in M$ Contradiction! Therefore We get, $2,3,5 \notin M$ and one of $(7,8,14)$ and one of $(1,4,9)$ not in $M \blacksquare$ From these We can construct $|M|=10$ elements cancelling $1,2,3,5,7$ $$M=(4,6,8,9,10,11,12,13,15) \blacksquare$$