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May 21st, 2019, 09:45 PM  #1 
Senior Member Joined: May 2015 From: Arlington, VA Posts: 458 Thanks: 29 Math Focus: Number theory  a+(b*c)=a*(b+c)
Does a+(b*c)=a*(b+c) have any solutions besides {0, 0, 0} or {1, 1, 1}?

May 21st, 2019, 10:21 PM  #2 
Member Joined: Oct 2018 From: USA Posts: 90 Thanks: 61 Math Focus: Algebraic Geometry 
$\{1,2,1\}$ $\{2,3,4\}$ EDIT: You can do some algebra to reach $\displaystyle a=\frac{bc}{b+c1}$ Let $c=b+1$, then $\displaystyle a=\frac{b(b+1)}{b+(b+1)1}$ $\displaystyle a=\frac{b+1}{2}$ So just choose some arbitrary odd $b$ and any set of the form $\displaystyle \left\{\frac{b+1}{2} \; , \; b \; , \; b+1 \right\}$ works. Last edited by Greens; May 21st, 2019 at 10:31 PM. Reason: Found Another, Pattern 
May 21st, 2019, 10:28 PM  #3 
Senior Member Joined: May 2015 From: Arlington, VA Posts: 458 Thanks: 29 Math Focus: Number theory 
Do you have a method for finding these solutions, Greens, and are they countless?

May 21st, 2019, 10:46 PM  #4 
Member Joined: Oct 2018 From: USA Posts: 90 Thanks: 61 Math Focus: Algebraic Geometry 
I saw your reply as I added my edit, bit of awkward timing. The edit should answer your question.

May 22nd, 2019, 03:43 PM  #5 
Math Team Joined: Oct 2011 From: Ottawa Ontario, Canada Posts: 14,597 Thanks: 1038 
25 solutions keeping a, b and c in range 1 to 9: 111, 112, 113, 114, 115, 116, 117, 118, 119, 121, 131, 141, 151, 161, 171, 181, 191, 234, 243, 349, 356, 365, 394, 478, 487. 