I am having trouble determining if my solution is a valid proof or not:

Suppose $a,b,c$ make up three consecutive terms in an arithmetic sequence. Prove that $a^2-bc, b^2-ac, c^2-ab$ are also terms in an arithmetic sequence.

The question is from an old math contest and this solution isn't one of the ones offered. Here is what I've done:

Since $a,b,c$ are consecutive terms in an arithmetic sequence there exists a number $k$ so that:

$$k=b-a=c-b$$

Next we define the numbers $r_1$ and $r_2$ so that:

$$r_1=(b^2-ac) - (a^2-bc)$$ $$r_2=(c^2-ab) - (b^2-ac)$$

If $r_1 = r_2$ then we know $a^2-bc, b^2-ac, c^2-ab$ form an arithmetic sequence.

Rearrange the first equation so that:

$$r_1 = (b+a)(b-a) +c(b-a)$$

and notice that $b-a=k$,so we can express $r_1$ as:

$$r_1=k(a+b+c)$$

Similarily for $r_2$:

$$r_2 = (c+b)(c-b)+a(c-b)$$

Noting once again that $c-b=k$; $r_2 = k(a+b+c)$. Therefore $r_1=r_2$. $\qquad\blacksquare$

Is this proof valid?

• Yes, it’s fine; rather nice, in fact. – Brian M. Scott Aug 7 '16 at 19:52
• Yeah it is!good proof – Sathasivam K Aug 7 '16 at 19:53
• $a^2 - bc = k^2 - 3 b k,$ $b^2 - ca = k^2,$ $c^2 - a b = k^2 + 3 b k$ – Will Jagy Aug 7 '16 at 19:56

Your proof is correct. However, it is a proof that contains a 'rabbit' (see the introduction of EWD1300): it introduces the expression $\;(b+a)(b−a)+c(b−a)\;$ as a complete surprise to the reader. That makes the proof harder to read, harder to find, and harder to remember and reconstruct later.

Here are two alternative proofs which avoid that pitfall: they solve this problem essentially by calculation.\newcommand{\calc}{\begin{align} \quad &} \newcommand{\op}[1]{\\ #1 \quad & \quad \unicode{x201c}} \newcommand{\hints}[1]{\mbox{#1} \\ \quad & \quad \phantom{\unicode{x201c}} } \newcommand{\hint}[1]{\mbox{#1} \unicode{x201d} \\ \quad & } \newcommand{\endcalc}{\end{align}} \newcommand{\Ref}[1]{\text{(#1)}} \newcommand{\then}{\Rightarrow} \newcommand{\when}{\Leftarrow} \newcommand{\true}{\text{true}} \newcommand{\false}{\text{false}}

As you observed, $\;x,y,z\;$ are an arithmetic sequence iff $\;y-x = z-y\;$, or simplified $\;2y = x+z\;$, or (solving for $\;y\;$) $\;y = \tfrac 1 2 (x+z)\;$.

That last form gives us a very straightforward proof: given $\;b = \tfrac 1 2 (a+c)\;$, you are asked to prove $$\tag 1 b^2-ac = \tfrac 1 2 ((a^2-bc)+(c^2-ab))$$ So we can just substitute $\;b := \tfrac 1 2 (a+c)\;$ in $\Ref 1$, and then prove the resulting identity by simplifying and calculation.

And using the form $\;2y = x+z\;$ gives another easy-to-find proof, which I find more appealing: asuming $\;2b = a+c\;$, we calculate $$\calc \tag{1'} 2(b^2-ac) = (a^2-bc)+(c^2-ab) \op\equiv\hint{move all negative terms to the other side} 2b^2 + bc + ab = a^2 + c^2 + 2ac \op\equiv\hints{simplify left hand side; simplify right hand side} \hints{-- this introduces both \;2b\; and \;a+c\;, which} \hint{we know something about} b(2b + c + a) = (a + c)^2 \op\equiv\hint{substitute \;2b\; for \;a+c\;, by the assumption} b(2b + 2b) = (2b)^2 \op\equiv\hint{arithmetic} \true \endcalc$$

Here's hoping that this gets more people interested in rabbit extermination:

We don’t want to baffle or puzzle our readers [...] As time went by, we accepted as challenges to avoid pulling rabbits out of the magician’s hat. [...] Eventually, expelling rabbits became another joy of my professional life.

-- Edsger W. Dijkstra, from "The notational conventions I adopted, and why" (EWD1300)

I know your point is not to get another solution.

Anyway, I would prefer to rewrite the terms with $b-r,b,b+r$, giving $$(b-r)^2-b(b+r),b^2-(b-r)(b+r),(b+r)^2-(b-r)b$$ or $$-3br+r^2,r^2,3br+r^2.$$