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- Feb 14, 2012

- 3,909

- Thread starter anemone
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- Feb 14, 2012

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- Feb 7, 2012

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I don't have a careful proof, but here is how I came across the solution (which I am fairly sure is unique). Write $\sum a$ for $a+b+c$ and $\sum bc$ for $bc+ca+ab$. Then $\sum a,\ \sum bc$ and $abc$ form an arithmetic progression, and therefore $\sum a - 2\sum bc + abc = 0.$ Hence $$\textstyle (a-2)(b-2)(c-2) = abc - 2\sum bc + 4 \sum a - 8 = 3\sum a - 8.$$

That equation tells you quite a lot. For a start, the product $(a-2)(b-2)(c-2)$ is congruent to $1\pmod3$, which means that

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- Feb 7, 2012

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Good catch: I completely missed that one.3,7,16 : 26,181,336 ; diff=155

Not allowed! Those numbers are not4,4,24 : 32,208,384 ; diff=176

True....but that was a BONUS: I DID state: (keeping it a =< b =< c)Not allowed! Those numbers are notdistinct.

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- Feb 14, 2012

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My solution:

By continuing with what Opalg has mentioned in his post

That equation tells you quite a lot. For a start, the product $(a-2)(b-2)(c-2)$ is congruent to $1\pmod3$, which means thateither$a$, $b$ and $c$ are all multiples of $3$,orone of them is a multiple of $3$ and the other two are congruent to $1\pmod3$. Next, the product of the three numbers $a-2$, $b-2$, $c-2$ is only about three times their sum, so at least one of them must be rather small. You can easily rule out the possibility that the smallest of the numbers $a,\ b,\ c$ is $1$. So that makes it seem almost sure to be $3$. If you then put $a=3$ in the displayed equation",

And the difference between terms is always the same for any arithmetic progression, we have

\(\displaystyle 3bc-(3(b+c)+bc)=3(b+c)+bc-(3+b+c)\)

\(\displaystyle bc=5b+5c-3\)

\(\displaystyle b=\frac{5c-3}{c-5}=5+\frac{22}{c-5}\)

Since we're told that both $b$ and $c$ are distinct positive integers, this leads to only three possible values of $c-5$ that it could take as $22=1(22)=2(11)$.

If \(\displaystyle c-5=22\;\;\rightarrow c=27,\;b=6,\;\;a=3\).

If \(\displaystyle c-5=2\;\;\rightarrow c=7,\;b=16,\;\;a=3\).

If \(\displaystyle c-5=11\;\;\rightarrow c=16,\;b=7,\;\;a=3\).

Hence, $(a, b, c)$=$(3, 6, 27)$ or $(3, 16, 7)$.