If you found it on the forum why you did't insert the link here? Anyway, we include in $A$ some number of 2 to obtain $\prod_{x\in A} x > \sum_{x\in A} x^2$. Let $A=\{x_1,x_2,...,x_k\}$. Consider the sequence $x_{n+1}=x_1x_2...x_n-1$, $n\geq k$. Denote $A_n=\{x_1,x_2,...,x_n\}$ ($n\geq k$). Then $A\subset A_n$. Let's calculate $s_{n+1}=\prod_{x\in A_{n+1}} x - \sum_{x\in A_{n+1}} x^2$. Indeed, using recurrent expression for $x_{n+1}$ we immediately obtain \[s_{n+1}=\prod_{x\in A_{n}} x - \sum_{x\in A_{n}} x^2-1=s_n-1.\] Taking into consideration $s_k>0$ we conclude there is an index $K$ s.t. $s_K=0$. That's all.

Myth wrote: If you found it on the forum why you did't insert the link here? Heh, it doesn't really help since there's no solution, but anyways it's here: http://www.artofproblemsolving.com/Forum/viewtopic.php?highlight=magnificent+NT+problem&t=8427 Just one question: what made you think of using $x_{n+1}=x_1x_2\cdots x_n -1$? (I liked your solution btw )

Myth wrote: That's all Well done!!! __________ blang

ThAzN1 wrote: Just one question: what made you think of using $x_{n+1}=x_1x_2\cdots x_n -1$? (I liked your solution btw ) It was an inner voice. Actually, I think, it was inspired by an old russian problem.

Do you remember the problem?

No, I don't.

Myth's argument has a small gap: A and B should be sets, whereas he is using several copies of the number 2.

test20 wrote: Myth's argument has a small gap: A and B should be sets, whereas he is using several copies of the number 2. It doesn't matter. I used it to obtain $\prod_{x\in A} x > \sum_{x\in A} x^2$. Indeed, we can do it in many ways.

Yeah, if we take $m=\text{max}\{x_k\}$ then we can add $m+1, m+2, \cdots$, and I think it'll work.

Task appeared also on the IMO training camp, Poland 1999.

I think this is a much more motivated solution By adding $M,M+1,M+2$ to $A$ for some massive $M$, we can WLOG assume that $\prod_{x \in A} x>\sum_{x \in A} x^2+1000$. Let $p=\prod_{x \in A} x$ and $q=\sum_{x \in A} x^2$; note that $p>1000$. Now let $n=p-q$; we will find $n$ distinct massive positive integers $x_1,\ldots,x_n$ (so that $x_i \not \in A$) such that $$x_1^2+\cdots+x_n^2-px_1\ldots x_n+q=0,$$whence $B=A \cup \{x_1,\ldots,x_n\}$ will work. This is by Vieta jumping, with the initial solution being $(1,\ldots,1)$. If at some point we have a solution where (WLOG) $x_1 \leq \cdots \leq x_n$. Fixing $x_2,\ldots,x_n$, we get a quadratic for $x_1$ with two integer roots; one of these is the current value of $x_1$, and the other is $px_2\ldots x_n-x_1>x_n$. If we repeat this process many times, each variable grows arbitrarily large, and furthermore, from the $n$th iteration onwards, all the $x_i$ will be distinct, so we are done. $\blacksquare$