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.


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 $$-3br+r^2,r^2,3br+r^2.$$