Sum of Pair of Elements of Geometric Sequence with Three Elements in Lowest Terms is Coprime to other Element

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Theorem

Let $P = \tuple {a, b, c}$ be a geometric sequence of integers in its lowest terms.

Then $\paren {a + b}$, $\paren {b + c}$ and $\paren {a + c}$ are all coprime to each of $a$, $b$ and $c$.


In the words of Euclid:

If three numbers in continued proportion be the least of those which have the same ratio with them, any two whatsoever added together will be prime to the remaining number.

(The Elements: Book $\text{IX}$: Proposition $15$)


Proof

Let the common ratio of $P$ in canonical form be $\dfrac q p$.

By Form of Geometric Sequence of Integers in Lowest Terms:

$P = \tuple {p^2, p q, q^2}$

Then:

\(\ds p\) \(\perp\) \(\ds q\) Definition of Canonical Form of Rational Number
\(\ds \leadsto \ \ \) \(\ds q\) \(\perp\) \(\ds \paren {p + q}\) Numbers are Coprime iff Sum is Coprime to Both
\(\ds \leadsto \ \ \) \(\ds q\) \(\perp\) \(\ds p \paren {p + q}\) Integer Coprime to all Factors is Coprime to Whole
\(\ds \leadsto \ \ \) \(\ds q^2\) \(\perp\) \(\ds p \paren {p + q}\) Square of Coprime Number is Coprime
\(\ds \leadsto \ \ \) \(\ds q^2\) \(\perp\) \(\ds p^2 + p q\) factorising
\(\ds \leadsto \ \ \) \(\ds c\) \(\perp\) \(\ds a + b\)


Similarly:

\(\ds \leadsto \ \ \) \(\ds p^2\) \(\perp\) \(\ds p q + q^2\)
\(\ds \leadsto \ \ \) \(\ds a\) \(\perp\) \(\ds b + c\) factorising


Then:

\(\ds p + q\) \(\perp\) \(\ds p\) Numbers are Coprime iff Sum is Coprime to Both
\(\, \ds \land \, \) \(\ds p + q\) \(\perp\) \(\ds q\) Numbers are Coprime iff Sum is Coprime to Both
\(\ds \leadsto \ \ \) \(\ds \paren {p + q}^2\) \(\perp\) \(\ds p q\) as above
\(\ds \leadsto \ \ \) \(\ds p^2 + q^2 + 2 p q\) \(\perp\) \(\ds p q\) factorising
\(\ds \leadsto \ \ \) \(\ds p^2 + q^2\) \(\perp\) \(\ds p q\)
\(\ds \leadsto \ \ \) \(\ds b\) \(\perp\) \(\ds a + c\)

$\blacksquare$


Historical Note

This proof is Proposition $15$ of Book $\text{IX}$ of Euclid's The Elements.


Sources