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Chapter 1: Problem 195

For \(n\) a positive integer, show that the number of integral solutions \((x, y)\) of \(x^2+x y+y^2=n\) is finite and a multiple of 6.

Short Answer

Expert verified
The number of integral solutions is finite and a multiple of 6.

Step by step solution

01

Understanding the Problem

Given a positive integer \(n\), we need to show that the number of integral solutions \((x, y)\) for the equation \(x^2 + xy + y^2 = n\) is finite and a multiple of 6.
02

Rewrite the Equation

Rewrite \(x^2 + xy + y^2 = n\) as \((x - \frac{y}{2})^2 + \frac{3y^2}{4} = n\). This shows that the left side must be non-negative and finite.
03

Bounding the Variables

To ensure \(x^2 + xy + y^2 = n\), both \(x\) and \(y\) must be bounded. Specifically, \(-\frac{\text{sqrt}(4n)}{2} \le y \le \frac{\text{sqrt}(4n)}{2}\) and \(-\text{sqrt}(3n)/2 - y^2/(2x) \le x \le \text{sqrt}(3n)/2 - y^2/(2x)\).
04

Consider Different Values and Pairs

By considering each pair \((x, y)\) within the derived bounds, substitute into the equation to check for valid integral results. This ensures each valid integral pair \((x, y)\) is finite.
05

Count Solutions

Each solution \((x,y)\) will have corresponding permutations due to symmetry (i.e., rotational and reflection symmetry), each contributing similarly for being a multiple of 6.
06

Conclude the Solution

Hence, because each \(n\) leads to finite integral solutions and each is adjusted by the 6-fold symmetry, it results in finite counts that are multiples of 6.

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Integral Solutions
Diophantine equations deal with finding integer solutions to polynomial equations. For the equation given, we are looking for all pairs of integers \(x, y\) that satisfy \(x^2 + xy + y^2 = n\). Since these solutions must be integers, we refer to them as integral solutions. This means both x and y must be whole numbers. One approach to solve this is by rewriting the equation in a more manageable form. By doing this, the problem often simplifies into a form where a limited range of values can be considered for each variable. Given the constraints and nature of the original equation, it's also clear that there won't be an infinite number of such pairs.
Bounding Variables
To find the integral solutions, we need to understand the limits within which our variables can vary. Bounding the variables helps with efficiency and ensures we are only examining feasible values. The rewritten equation, \( (x - \frac{y}{2})^2 + \frac{3y^2}{4} = n \), tells us that both sides must remain positive and within a finite range. This gives us natural boundaries:
  • \ ( - \frac{\text{sqrt}(4n)}{2} \le y \le \frac{\text{sqrt}(4n)}{2} ) \
  • \ ( - \frac{\text{sqrt}(3n)}{2} - \frac{y^2}{2x} \le x \le \frac{\text{sqrt}(3n)}{2} - \frac{y^2}{2x} ) \
By substituting the values within these bounds into the original equation, we can identify the valid integral pairs \(x, y\). The key is to ensure that the calculated pairs remain finite and bounded by these ranges.
Symmetry in Equations
Symmetry plays an important role in simplifying and understanding the structure of Diophantine equations. In our given problem, the equation \(x^2 + xy + y^2 = n\) exhibits both rotational and reflection symmetry. This means that for any given solution pair \( (x,y) \), there exist equivalent pairs when we consider the symmetrical nature of the equation:
  • Reflecting over the x-axis and y-axis
  • Rotating around the origin
Due to these symmetries, each solution found has exactly 6 distinct configurations or permutations. Thus, the total number of integral solutions is not just finite, but is also a multiple of 6. This property hugely benefits us in counting and categorizing solutions, reflecting the inherent balance in the equation鈥檚 structure.

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Most popular questions from this chapter

Let \(p(x)=x^3+a_1 x^2+a_2 x+a_3\) have rational coefficients and have roots \(r_1, r_2, r_3\). If \(r_1-r_2\) is rational, must \(r_1, r_2\), and \(r_3\) be rational?

Consider the transformation of the plane (except for the coordinate axes) defined by sending the point \((x, y)\) to the point \((y+1 / x, x+1 / y)\). Suppose we apply this transformation repeatedly, starting with some specific point \(\left(x_0, y_0\right)\), to get a sequence of points \(\left(x_n, y_n\right)\). a. Show that if \(\left(x_0, y_0\right)\) is in the first or the third quadrant, the sequence of points will tend to infinity. b. Show that if \(\left(x_0, y_0\right)\) is in the second or the fourth quadrant, either the sequence will terminate because it lands at the origin, or the sequence will be eventually periodic with period 1 or 2 , or there will be infinitely many \(n\) for which \(\left(x_n, y_n\right)\) is further from the origin than \(\left(x_{n-1}, y_{n-1}\right)\).

Given 64 points in the plane which are positioned so that 2001, but no more, distinct lines can be drawn through pairs of points, prove that at least four of the points are collinear.

Find all twice continuously differentiable functions \(f\) for which there exists a constant \(c\) such that, for all real numbers \(a\) and \(b\), $$ \left|\int_a^b f(x) d x-\frac{b-a}{2}(f(b)+f(a))\right| \leq c(b-a)^4 $$

The other day, in the honors calculus class at Wohascum College, the instructor asked the students to compute \(\int_0^{\infty} \frac{e^{-x}-e^{-2 x}}{x} d x\). One student split up the integral and made the substitution \(u=2 x\) in the second part, concluding that $$ \int_0^{\infty} \frac{e^{-x}-e^{-2 x}}{x} d x=\int_0^{\infty} \frac{e^{-x}}{x} d x-\int_0^{\infty} \frac{e^{-u}}{u} d u=0 $$ The instructor was not too impressed by this, pointing out that for all positive values of \(x, \frac{e^{-x}-e^{-2 x}}{x}\) is positive, so how could the integral be zero? a. Resolve this paradox. b. Eventually a student gave up and asked Mathematica to compute the integral, and an exact answer appeared on the screen: \(\log [2]\) (which is Mathematica's notation for \(\ln 2\) ). Is this answer correct?
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