My Math Forum a+(b*c)=a*(b+c)

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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 $\displaystyle a=\frac{bc}{b+c-1}$ 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. Thanks from topsquark and Loren 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.

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