[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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#3
Wilmer
303
0
2 more, Opal (keeping it a =< b =< c):
3,7,16 : 26,181,336 ; diff=155
4,4,24 : 32,208,384 ; diff=176
#4
Opalg
Gold Member
MHB
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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.
#5
Wilmer
303
0
Opalg said:
Not allowed! Those numbers are not distinct.
True...but that was a BONUS: I DID state: (keeping it a =< b =< c)(Ninja)
#6
anemone
Gold Member
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POTW Director
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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 $22=1(22)=2(11)$.
Insights auto threads is broken atm, so I'm manually creating these for new Insight articles.
In Dirac’s Principles of Quantum Mechanics published in 1930 he introduced a “convenient notation” he referred to as a “delta function” which he treated as a continuum analog to the discrete Kronecker delta. The Kronecker delta is simply the indexed components of the identity operator in matrix algebra
Source: https://www.physicsforums.com/insights/what-exactly-is-diracs-delta-function/
by...
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$$
a^n+b^n=c^n
$$
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I'm interested to know whether the equation
$$1 = 2 - \frac{1}{2 - \frac{1}{2 - \cdots}}$$
is true or not. It can be shown easily that if the continued fraction converges, it cannot converge to anything else than 1. It seems that if the continued fraction converges, the convergence is very slow. The apparent slowness of the convergence makes it difficult to estimate the presence of true convergence numerically. At the moment I don't know whether this converges or not.
