Math Problem Statement

How to solve diophantine equation bc = 2(b^2+c^2-a^2) ?

Solution

To solve the given Diophantine equation bc=2(b2+c2a2)b \cdot c = 2(b^2 + c^2 - a^2), let's break it down systematically.

Equation:

bc=2(b2+c2a2)b \cdot c = 2(b^2 + c^2 - a^2)

Rearranging terms:

Expand and simplify: bc=2b2+2c22a2b \cdot c = 2b^2 + 2c^2 - 2a^2 bc2b22c2+2a2=0b \cdot c - 2b^2 - 2c^2 + 2a^2 = 0 Factor where possible: bc2(b2+c2)+2a2=0b \cdot c - 2(b^2 + c^2) + 2a^2 = 0

Analyzing the equation:

This is a quadratic Diophantine equation in integers b,c,b, c, and aa. To find solutions, follow these steps:


Step 1: Isolate terms and explore factorization

Reorganize: bc=2b2+2c22a2b \cdot c = 2b^2 + 2c^2 - 2a^2 Divide through by 2 (to simplify if nonzero): bc2=b2+c2a2\frac{b \cdot c}{2} = b^2 + c^2 - a^2

Let us rewrite this equation to look for special patterns.


Step 2: Check simple cases

Set b,c,ab, c, a to specific values and verify integer solutions.

  1. Try small integers b=c=1b = c = 1: 11=2(12+12a2)    1=2(2a2)1 \cdot 1 = 2(1^2 + 1^2 - a^2) \implies 1 = 2(2 - a^2) 1=42a2    2a2=3(no integer solutions).1 = 4 - 2a^2 \implies 2a^2 = 3 \quad (\text{no integer solutions}).

  2. Try b=2,c=1b = 2, c = 1: 21=2(22+12a2)    2=2(4+1a2)2 \cdot 1 = 2(2^2 + 1^2 - a^2) \implies 2 = 2(4 + 1 - a^2) 2=102a2    2a2=8    a2=4    a=2.2 = 10 - 2a^2 \implies 2a^2 = 8 \implies a^2 = 4 \implies a = 2.

Thus, one solution is (b,c,a)=(2,1,2)(b, c, a) = (2, 1, 2).


Step 3: Explore general forms

Factorization methods and trial solutions can reveal other integer solutions. Rewrite: bc=2(b2+c2a2)b \cdot c = 2(b^2 + c^2 - a^2) bc=2b2+2c22a2b \cdot c = 2b^2 + 2c^2 - 2a^2

Introduce auxiliary variables, substitutions, or parameterizations for systematic exploration.


Solutions:

The solution space can often be parametrized. Integer solutions like (b,c,a)=(2,1,2)(b, c, a) = (2, 1, 2), or others found numerically or algebraically, are possible. Further algebraic techniques, such as completing the square or modular arithmetic, may refine the search.


Would you like me to expand on specific techniques or explore further solutions? Let me know!


Expanding your understanding:

  1. How can completing the square help solve quadratic Diophantine equations?
  2. How do modular arithmetic constraints narrow down possible solutions?
  3. Can this equation be linked to geometric interpretations, like Pythagorean triples?
  4. What is the significance of parameterizing solutions in two variables?
  5. How does trial and error combined with factorization assist in finding integer solutions?

Tip:

Always check for symmetries in a Diophantine equation to reduce the computational effort for solving them.

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Math Problem Analysis

Mathematical Concepts

Diophantine Equations
Algebra
Quadratic Equations
Factorization
Integer Solutions

Formulas

b*c = 2(b^2 + c^2 - a^2)
Rearranged: b*c - 2(b^2 + c^2) + 2a^2 = 0

Theorems

Fundamental theorem of arithmetic (used implicitly for factorization)
Quadratic solution techniques for integer solutions

Suitable Grade Level

Undergraduate or advanced high school (Grades 11-12)