A Mathematical Rabbit out of an Algebraic Hat

This is how G. D. Chakerian has described his derivation of the 3-term Arithmetic Mean - Geometric Mean Inequality (also Arithmetic-Geometric Mean Inequality or simply AM-GM Inequality) in a chapter of the Mathematical Plums (edited by R. Honsberger.) The inequality asserts that

For any non-negative real numbers $a, b, c,$

(1)

$\sqrt[3]{abc} \le (a + b + c) / 3,$

with equality only if $a = b = c.$

The proof depends on the following identity that holds for non-negative $x, y, z$:

(2)

$\begin{align} x^{3} + y^{3} &+ z^{3} - 3xyz\\ &= \frac{1}{2}(x + y + z)((x - y)^{2} + (y - z)^{2} + (z - x)^{2}) \end{align}$

which is just another form for a more obvious

(3)

$\begin{align} x^{3} + y^{3} + z^{3} &- 3xyz\\ &= (x + y + z)(x^{2} + y^{2} + z^{2} - xy - yz - zx) \end{align}$

The right-hand side of the identity in (2) is not negative and so is the left-hand side, implying

$x^{3} + y^{3} + z^{3} \ge 3xyz.$

Now simply substitute $a = x^{3},$ $b = y^{3},$ and $c = z^{3}$ to obtain the AM-GM inequality (1).


Now, for any one who does little mathematics and is not accustomed to problem solving, the derivation, although simple, does appear as a trick out of the blue. Certainly it takes knowledge and experience to come with such a proof. In fact, identity (3) is more basic than (2). The inequality

$x^{2} + y^{2} + z^{2} - xy - yz - zx \ge 0$

can be obtained as the sum of three simpler inequalities

$\begin{align} x^{2} + y^{2} &\ge 2xy,\\ y^{2} + z^{2} &\ge 2yz,\\ z^{2} + x^{2} &\ge 2zx, \end{align}$

which of course explains (2) and might make it less of a mystery.

The identity (2) and similar ones do make frequent appearances in a multitude of settings and are not too difficult to remember.

Problem

(Andreescu et al, p. 9.)

Find all triples $(x, y, z)$ of positive integers such that

$x^{3} + y^{3} + z^{3} - 3xyz = p$

where $p$ is a prime greater than $3.$

The solution comes from (3). Since $p$ is prime and $x + y + z \ge 1$ it must be that $x + y + z = p$ and $x^{2} + y^{2} + z^{2} - xy - yz - zx = 1.$ Now (2) implies that

(4)

$(x - y)^{2} + (y - z)^{2} + (z - x)^{2} = 2.$

Without loss of generality, we may assume $x \ge y \ge z.$ Then from (4), it is either $x = y = z + 1$ or $x - 1 = y = z.$ Prime $p$ has one of the forms $3k + 1$ or $3k + 2.$ In the first case the solutions are $((p+2)/3, (p-1)/3, (p-1)/3);$ in the second case the solutions are $((p+1)/3, (p+1)/3, (p-2)/3).$ For a complete list of solutions, the triples need to be permuted.

References

  1. R. Honsberger, Mathematical Plums, MAA, 1979
  2. T. Andreescu et al, An Introduction to Diophantine Equations, Birkhäuser, 2010

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