MHB Arithmetic Progression Challenge

[anemone]

Find distinct positive integers $$a,\;b,$$ and $$c$$ such that $$a+b+c,\;ab+bc+ac,\;abc$$ forms an arithmetic progression.

 

[Opalg]

[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]

 

[Wilmer]

2 more, Opal (keeping it a =< b =< c):

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

4,4,24 : 32,208,384 ; diff=176

 

[Opalg]

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

Good catch: I completely missed that one.

4,4,24 : 32,208,384 ; diff=176

Not allowed! Those numbers are not distinct.

 

[Wilmer]

Not allowed! Those numbers are not distinct.

True...but that was a BONUS: I DID state: (keeping it a =< b =< c)(Ninja)

 

[anemone]

Thank you Opalg and Wilmer for participating...

My solution:

By continuing with what Opalg has mentioned in his post

Spoiler

"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 $(3, 16, 7)$.