Q. \(a^2 + b^2 = c^2\).

Solution and detailed explanation

We are given the Diophantine equation

\(a^2 + b^2 = c^2\).

We will derive the complete family of integer solutions (the Pythagorean triples) and give conditions for primitive triples. The argument proceeds by converting the equation to a rational point on the unit circle and parameterizing those rational points. The derivation is given step by step with full explanations.

1. Normalize by c (divide both sides by c^2).

   Assume c is nonzero. Define the ratios \(x = \dfrac{a}{c}\) and \(y = \dfrac{b}{c}\). Then the equation becomes

   \(x^2 + y^2 = 1\).

   We are now looking for rational points \((x,y)\) on the unit circle, because if \(a,b,c\) are integers then \(x,y\) are rational numbers.

2. Parameterize rational points on the unit circle using a rational slope.

   Take a fixed rational parameter \(t\). Consider the line through the point \((-1,0)\) on the unit circle with slope \(t\). The equation of that line is

   \(y = t(x + 1)\).

   Intersect this line with the circle \(x^2 + y^2 = 1\). Substitute \(y\) from the line into the circle:

   \(x^2 + t^2(x + 1)^2 = 1\).

   Expand and collect terms:

   \((1 + t^2)x^2 + 2t^2 x + (t^2 – 1) = 0\).

   One root of this quadratic is \(x = -1\) (the known intersection), so factor to find the other root. Solving yields the other intersection

   \(x = \dfrac{1 – t^2}{1 + t^2}\),

   and then

   \(y = \dfrac{2t}{1 + t^2}\).

   Thus every rational slope \(t\) produces a rational point on the unit circle. Conversely, every rational point except \((-1,0)\) arises this way for some rational \(t\).

3. Return to integer variables.

   Let the rational parameter \(t\) be expressed as a fraction of integers, \(t = \dfrac{n}{m}\), where \(m\) and \(n\) are integers with \(m \neq 0\). Substitute into the parameterization formulas and clear denominators by multiplying numerator and denominator by \(m^2\):

   \(x = \dfrac{m^2 – n^2}{m^2 + n^2}\),

   \(y = \dfrac{2mn}{m^2 + n^2}\).

   Recall \(x = \dfrac{a}{c}\) and \(y = \dfrac{b}{c}\). Thus one can take

   \(\displaystyle a = k(m^2 – n^2),\)

   \(\displaystyle b = k(2mn),\)

   \(\displaystyle c = k(m^2 + n^2),\)

   for some positive integer scaling factor \(k\). Substituting these into \(a^2 + b^2 = c^2\) verifies the identity because

   \((m^2 – n^2)^2 + (2mn)^2 = (m^2 + n^2)^2.\)

4. Check the algebraic identity (verification).

   Compute the left-hand side:

   \((m^2 – n^2)^2 + (2mn)^2 = m^4 – 2m^2n^2 + n^4 + 4m^2n^2 = m^4 + 2m^2n^2 + n^4 = (m^2 + n^2)^2.\)

   Multiplying each of \(a,b,c\) by the same integer \(k\) preserves the equation, so the formulas with factor \(k\) produce all integer solutions up to ordering of \(a\) and \(b\).

5. Primitive triples and parity/coprimality conditions.

   An integer triple \((a,b,c)\) is primitive if gcd(a,b,c) = 1. The parameterization above produces a primitive triple when one chooses integers \(m,n\) satisfying the following conditions:

   • m and n are coprime, gcd(m,n) = 1,
   • m and n have opposite parity: one is even and the other is odd,
   • m > n > 0.

   Under these conditions and taking k = 1, the triple

   \((a,b,c) = (m^2 – n^2,\; 2mn,\; m^2 + n^2)\)

   is a primitive Pythagorean triple. If the parity condition is not satisfied or m and n are not coprime, the triple may be non-primitive; in that case choose k > 1 or reduce by the common divisor.

6. Summary: the full family of integer solutions.

   All integer solutions to \(a^2 + b^2 = c^2\) are given, up to swapping a and b and up to an overall sign, by the formulas

   \(\displaystyle a = k\,(m^2 – n^2),\)

   \(\displaystyle b = k\,(2mn),\)

   \(\displaystyle c = k\,(m^2 + n^2),\)

   where \(m\) and \(n\) are positive integers with \(m > n\), \(k\) is a positive integer, and for primitive triples one additionally requires gcd(m,n) = 1 and that m and n have opposite parity.

This completes the step-by-step derivation and explanation of all integer solutions to a square plus b square equals c square.