Show that the cube roots of three distinct prime numbers cannot be three terms (not necessarily consecutive) of an arithmetic progression.
Problem
Source: 1973 USAMO Problem 5
Tags: geometry, 3D geometry, number theory, prime numbers, arithmetic sequence
m.candales
07.03.2010 06:10
Suppose the contrary; and let $ \sqrt {p}, \sqrt {q}, \sqrt {r}$ be three distinct terms of an arithmetic progression, where $ p, q, r$ are primes. Then $ \sqrt {p} - \sqrt{q} = md$ and $ \sqrt {q} - \sqrt {r} = nd$ where $ m, n \in Z$ and $ d$ is the common difference. Then $ \frac {\sqrt {p} - \sqrt {q}}{\sqrt {q} - \sqrt {r}} = \frac {m}{n} = k \in Q$ $ \Longrightarrow \sqrt {p} + k\sqrt {r} = (k + 1)\sqrt {q}$ $ \Longrightarrow p + 2k\sqrt {pr} + k^2r = (k + 1)^2q$ $ \Longrightarrow \sqrt {pr} = \frac {(k + 1)^2q - p - k^2r}{2k} \in Q$ Which is a contradiction, because $ pr$ is not a perfect square, and then $ \sqrt{pr}$ is irrational
varunrocks
07.03.2010 06:44
Your solution might be right for square roots but is not true for a cube root which is clearly said in the problem.
futuremospper
07.03.2010 19:34
The solution is nearly identical...
Let $ \sqrt[3]{p}=a$, $ \sqrt[3]{q}=a+md$, and $ \sqrt[3]{r}=a+nd$, where $ p,q,r$ are distinct primes and $ m,n$ are integers. Eliminating $ a$ and $ d$ yields $ \frac{\sqrt[3]{q}-\sqrt[3]{p}}{\sqrt[3]{r}-\sqrt[3]{p}} = \frac{m}{n}$, or
\[ m\sqrt[3]{r} - n\sqrt[3]{q} = (m-n)\sqrt[3]{p}.\]
Cubing this yields
\[ m^3r - n^3q - 3mn \sqrt[3]{rq} (m\sqrt[3]{r}-n\sqrt[3]{q})=(m-n)^3p,\]
so
\[ 3mn(m-n)\sqrt[3]{pqr}=m^3r-n^3q-(m-n)^3p.\]
But $ \sqrt[3]{pqr}$ is irrational; hence, the two sides cannot be equal, and we have the desired contradiction.
professordad
30.03.2012 02:02
We prove by contradiction. Assume that there do exist such numbers $p_1$, $p_2$, and $p_3$ such that $\sqrt[3]{p_1}$, $\sqrt[3]{p_2}$, $\sqrt[3]{p_3}$ form an arithmetic progression. Then, $\sqrt[3]{p_1} + ad = \sqrt[3]{p_2}$ and $\sqrt[3]{p_2} + bd = \sqrt[3]{p_3}$ for common difference $d$ and integers $a$, $b$. Cubing gives
\begin{align*}p_1 + (ad)^3 + 3ad\sqrt[3]{p_1} \cdot (\sqrt[3]{p_1} + ad) &= p_2\\ p_2 + (bd)^3 + 3bd\sqrt[3]{p_2} \cdot (\sqrt[3]{p_2} + bd) &= p_3\end{align*} Plugging in $\sqrt[3]{p_1} + ad = p_2$ and $\sqrt[3]{p_2} + bd = p_3$,
\begin{align*}3ad\sqrt[3]{p_1} \cdot (\sqrt[3]{p_2}) &= \text{integer}\\ 3bd\sqrt[3]{p_2} \cdot (\sqrt[3]{p_3}) &= \text{integer}\end{align*} Dividing gives
\[\frac{a}{b} \cdot \sqrt[3]{\frac{p_1}{p_3}} = \text{rational} \Longrightarrow \sqrt[3]{\frac{p_1}{p_3}} = \text{rational}\] But this is impossible because $p_1$ and $p_3$ don't share any factors besides 1, and no prime numbers are perfect cubes. Contradiction.
Abra_Kadabra
08.04.2020 04:12
@above I think your solution has a flaw in it. You assumed that d, the common difference, is an integer, but that need not be the case. Also you have a typo, you said that: cuberoot(p_1) + ad = p_2 and cuberoot(p_2) + bd = p_3, (this minor typo does not affect the solution) it should really be: cuberoot(p_1) + ad = cuberoot(p_2) and cuberoot(p_2) + bd = cuberoot(p_3)
huashiliao2020
13.04.2023 07:11
Let $ \sqrt[3]{p}=a$, $ \sqrt[3]{q}=a+md$, and $ \sqrt[3]{r}=a+nd$, where $ p,q,r$ are distinct primes and $ m,n$ are integers. $ \frac{\sqrt[3]{q}-\sqrt[3]{p}}{\sqrt[3]{r}-\sqrt[3]{p}} = \frac{m}{n}$, or \[ m\sqrt[3]{r} - n\sqrt[3]{q} = (m-n)\sqrt[3]{p}.\] Cubing this yields \[ m^3r - n^3q - 3mn \sqrt[3]{rq} (m\sqrt[3]{r}-n\sqrt[3]{q})=(m-n)^3p,\]but the LHS has an irrational number, namely $(rq)^{1/3}(mr^{1/3}-nq^{1/3})$, since r^2q and rq^2 are not perfect cubes.