MHB Arithmetic Progression Challenge

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Distinct positive integers \(a\), \(b\), and \(c\) can be found such that \(a+b+c\), \(ab+bc+ac\), and \(abc\) form an arithmetic progression, with the solutions being \((3, 6, 27)\) and \((3, 7, 16)\). The arithmetic progression condition leads to the equation \((a-2)(b-2)(c-2) = 3\sum a - 8\), which provides insights into the values of \(a\), \(b\), and \(c\). The analysis shows that at least one of the integers must be small, leading to the conclusion that \(a\) is likely \(3\). The distinct integers must satisfy the conditions derived from the arithmetic progression, confirming the validity of the solutions.
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Find distinct positive integers $$a,\;b,$$ and $$c$$ such that $$a+b+c,\;ab+bc+ac,\;abc$$ forms an arithmetic progression.
 
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[sp]If $(a,b,c) = (3,6,27)$ then $a+b+c = 36$, $bc+ca+ab = 261$ and $abc = 486$. These numbers are in arithmetic progression.

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 either $a$, $b$ and $c$ are all multiples of $3$, or one 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 use the fact the other two numbers must either both be multiples of $3$ or both congruent to $1\pmod3$, it does not take long to find the solution.[/sp]
 
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2 more, Opal (keeping it a =< b =< c):

3,7,16 : 26,181,336 ; diff=155

4,4,24 : 32,208,384 ; diff=176
 
Wilmer said:
3,7,16 : 26,181,336 ; diff=155
Good catch: I completely missed that one.

Wilmer said:
4,4,24 : 32,208,384 ; diff=176
Not allowed! Those numbers are not distinct.
 
Opalg said:
Not allowed! Those numbers are not distinct.
True...but that was a BONUS: I DID state: (keeping it a =< b =< c)(Ninja)
 
Thank you Opalg and Wilmer for participating...

My solution:

By continuing with what Opalg has mentioned in his post

Opalg said:
"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 either $a$, $b$ and $c$ are all multiples of $3$, or one 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",

we could then rewrite the three consecutive terms of the arithmetic progression as $$3+b+c,\;\;3(b+c)+bc,\;\;3bc$$.

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

$$3bc-(3(b+c)+bc)=3(b+c)+bc-(3+b+c)$$

$$bc=5b+5c-3$$

$$b=\frac{5c-3}{c-5}=5+\frac{22}{c-5}$$

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 $$c-5=22\;\;\rightarrow c=27,\;b=6,\;\;a=3$$.

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

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

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