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May 21st, 2019, 09:45 PM   #1
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a+(b*c)=a*(b+c)

Does a+(b*c)=a*(b+c) have any solutions besides {0, 0, 0} or {1, 1, 1}?
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May 21st, 2019, 10:21 PM   #2
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$\{1,2,1\}$

$\{2,3,4\}$

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
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May 21st, 2019, 10:28 PM   #3
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Do you have a method for finding these solutions, Greens, and are they countless?
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May 21st, 2019, 10:46 PM   #4
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I saw your reply as I added my edit, bit of awkward timing. The edit should answer your question.
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May 22nd, 2019, 03:43 PM   #5
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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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