We present three theorems involving rational numbers.

## \(n\)th Power Theorem. Let \( n\) be a positive integer. If \( a\) is a positive integer such that \( a = r^n\) for some rational number \( r\), then \( r\) must be an integer.

Proof: We apply the integer root theorem to the polynomial \( x^n - a \). Since this polynomial has a rational root \( r\), this root must be an integer.\( _\square \)

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## Theorem 2. If the sum and product of two rational numbers are both integers, then the two rational numbers must be integers.

Proof: We have \( r_1 + r_2 = b\) and \( r_1 \cdot r_2 = c\), hence \( r_1, r_2\) are rational roots of the polynomial \( x^2 - bx + c\). By the integer root theorem, \( r_1\) and \( r_2\) are both integers. \( _\square \)

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## Theorem 3. A quadratic polynomial \( f(x) = ax^2 + bx +c \) has integer coefficients. Then both of the roots are integers if and only if:

## 1. \( b^2 - 4ac\) is a perfect square,

## 2. \( a\) divides both \( b\) and \( c\).

Proof: By the Remainder Factor Theorem, if the roots are integers \( n\) and \( m\), then \( f(x) = a (x-n)(x-m) \). By comparing terms, we obtain \( b = -a(n+m)\) and \( c = anm\). Hence, (2) is satisfied. Furthermore, \[ b^2 - 4ac = a^2 (n+m)^2 - 4 a \times anm = a^2 (n-m)^2,\] so (1) is also satisfied.

Conversely, by the quadratic formula, the roots of \(f(x)\) are \( \frac {-b \pm \sqrt{b^2- 4ac} }{2a} \). By condition (1), it follows that the roots are rational. By condition (2) and Vieta's formula, both the sum and product of the roots are integers. Hence by Theorem 2, the roots are integers. \( _\square \)

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TopNewestwhat's  ? – Daniel Lim · 2 years, 2 months ago

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– Calvin Lin Staff · 2 years, 2 months ago

It was meant to be a non-breaking space, to separate out indented paragraphs. I wanted the theorems to be displayed individually, instead of in a long chunk of text.Log in to reply

Edit: Actually, it's – Sean Ty · 2 years, 2 months ago

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