January 1st, 2017, 02:10 PM  #11 
Member Joined: Dec 2016 From: USA Posts: 46 Thanks: 11 
Note  I have no issue with your proof of the converse (i.e., that the product of two elements of $R$ can also be expressed as a sum of two elements of $R$).

January 1st, 2017, 02:13 PM  #12 
Member Joined: Dec 2016 From: USA Posts: 46 Thanks: 11  
January 1st, 2017, 03:01 PM  #13 
Member Joined: Dec 2016 From: USA Posts: 46 Thanks: 11 
@Rachel1234: These aren't normal HW problems. They are more like competition problems. My guess is that these problems were offered as challenge problems for extra credit, and that you are not required (and probably not expected) to submit solutions. Right? 
January 1st, 2017, 03:09 PM  #14 
Senior Member Joined: Sep 2015 From: USA Posts: 1,860 Thanks: 967  
January 1st, 2017, 03:20 PM  #15  
Senior Member Joined: Sep 2015 From: USA Posts: 1,860 Thanks: 967  Quote:
the required product is $A = \left(\sqrt{a b}+\dfrac{1}{\sqrt{a b}}\right)\left(\sqrt{\dfrac{a}{b}}+\sqrt{\dfrac{b }{a}}\right)$ and there is no guarantee that $\sqrt{a b},~\sqrt{\dfrac{a}{b}} \in \mathbb{Q}^+$ addendum: we have to factor in the info that $A \in \mathbb{Z}^+$, I'll continue to look at it Last edited by romsek; January 1st, 2017 at 03:23 PM.  
January 1st, 2017, 03:24 PM  #16 
Member Joined: Dec 2016 From: USA Posts: 46 Thanks: 11 
The statement is true  I can prove it.

January 1st, 2017, 03:56 PM  #17 
Senior Member Joined: Sep 2015 From: USA Posts: 1,860 Thanks: 967  
January 1st, 2017, 04:45 PM  #18 
Member Joined: Dec 2016 From: USA Posts: 46 Thanks: 11  The proof is elementary, but not a quick post  no time right now. Besides, my guess is that the problem is an extra credit problem for Rachel's class, so I'm not sure posting a full solution is warranted. I also have an elementary proof of part (B): "(B) Show that there are infinitely many positive integers $A$ that cannot be written as the sum of two elements of $R$." But I don't have a proof of part (C): "(C) Show that there are infinitely many positive integers $A$ that can be written as the sum of two elements of $R$." In fact, unless I'm missing something, part (C) is quite hard, and will probably require advanced methods. 
January 1st, 2017, 05:40 PM  #19 
Member Joined: Dec 2016 From: USA Posts: 46 Thanks: 11 
Ok, Rachel made it clear that the problems were from her math team class, assigned for fun, not for credit, so when I get a chance (sometime this week), I'll post my proofs for parts (A) and (B) (unless someone posts the same proof or a better one before me).

January 2nd, 2017, 08:49 AM  #20  
Senior Member Joined: Mar 2015 From: New Jersey Posts: 1,329 Thanks: 94  Quote:
Let $\displaystyle a_{ij}$ be a typical grid element. Then $\displaystyle a_{ij}=a_{ik}+a_{mj}, \ k\neq i, m\neq j$ EDIT: Whoops. Just saw this as a separate problem. Replied to that too. Last edited by zylo; January 2nd, 2017 at 08:54 AM.  

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