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\centerline{\bf What Else Pythagoras Could Have Done}
\centerline {Robert W. Floyd}
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Editor:

Yoram Sagher [See reference] shows that Pythagoras could have demonstrated
that the square root of an integer is an irrational or an integer without
knowing anything about prime numbers.  Assuming Pythagoras understood
Euclid's algorithm, the following proofs show how he could have demonstrated
that any integer root of an integer is an irrational or an integer, and even
that the cube root of an integer either is not the root of a quadratic (i.e. not
of the form $(a+b\ \sqrt n)/c)$ or is an integer.

In the following, all variables but $r$ are restricted to integer values.

The (shorter) proof of Sagher's special case comes first, to motivate the others.
If $r = \sqrt k = m/n$ (in lowest terms), Euclid's algorithm gives $\alpha$ and
$\beta$ for which $\alpha m+ \beta n = 1$.  Then $rn=m$, $rm=r↑2n=kn$, and
$r=r(\alpha m + \beta n) = \alpha rm + \beta rn = \alpha kn + \beta m$, an
integer.

Now take $r = m/n = \root 3 \of k$, $\alpha m + \beta n = 1$.  Then

$$m = rn,\ rn↑2 = mn,\ rmn = m↑2,\ rm↑2 = r↑3 n↑2 = kn↑2$$

$$r = r (\alpha m + \beta n)↑2 = \alpha ↑2 rm↑2 + 2 \alpha \beta rmn +
\beta ↑2 rn↑2 = \alpha ↑2 kn↑2 + 2 \alpha \beta m↑2 + \beta ↑2 mn, \hbox 
{ an integer.}$$

By an obvious generalization, for {\it any} integer $t \geq 2$, if $\root t \of k$
is rational it is an integer.

Now make the weaker assumption that 
 $r = \root 3 \of k$, and that $r$ satisfies a proper quadratic equation
$ar↑2 + br + c = 0$, $a ≠ 0$.  Then $0 = (ar↑2 + br + c) (ar - b) = a↑2 r↑3 +
(ac - b↑2) r - bc$.  If $ac = b↑2$, divide the equation by $a↑2$ to find $r↑3 =
(b/a)↑3$ and $r = b/a$.  
Otherwise, put $k$ for $r↑3$ and find $r = (bc - a↑2k)/(ac - b↑2)$.
Either way, $r$ is rational, and consequently an integer.

{\bf Reference}

Y. Sagher, What Pythagoras Could Have Done, American Mathematical Monthly,
\underbar{95} (1988) 117.
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