Perfect Squares in Pythagorean Triples: A Number-Theoretic Restriction

Number Theory

1. Divisibility
- 1.1. Greatest Common Divisor
  - 1.1.1. Important Problems On Greatest Common Divisor
- 1.2. Expressing Cubes As Differences Of Squares
2. Prime Numbers
- 2.1. Important Problems On Prime Numbers
- 2.2. Prime Number Theorem
  - 2.2.1. The Prime Number Theorem Explained With Interpretation
3. Congruence
- 3.1. Linear Congruence
  - 3.1.1. Important Applications On Linear Congruence
4. Number Theoretic Functions
- 4.1. Greatest Integer Functions
  - 4.1.1. Divisibility Of Binomial Coefficients
5. Quadratic Reciprocity
- 5.1. Legendre Symbol
  - 5.1.1. Legendre Symbol And Its Properties
6. Fermats Last Theorem
- 6.1. Proof Of Fermats Last Theorem
- 6.2. Pythagorean Triples
  - 6.2.1. Important Problems On Pythagorean Triples
7. Integers As Sum Of Squares
- 7.1. Sum Of More Than Two Squares
  - 7.1.1. Representation Of Integers As Sums Of Five Nonzero Squares For All Values Greater Than Or Equal To 170
- 7.2. Sum Of Two Squares
  - 7.2.1. Perfect Squares In Pythagorean Triples
8. Encryption And Decryption
- 8.1. Linear Cipher
  - 8.1.1. Important Applications On Linear Cipher
- 8.2. Hill Cipher
  - 8.2.1. Hill Cipher Decryption In Number Theory Step By Step With Example

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Perfect Squares in Pythagorean Triples

Explore the restriction of Perfect Squares in Pythagorean Triples. Understand key insights, patterns, and number-theoretic rules behind these special sets.

In a Pythagorean triple \(x, y, z\), not more than one of \(x\), \(y\), or \(z\) can be a perfect square.

Let \(x, y, z \in \mathbb{Z}^+\) be a primitive Pythagorean triple, i.e., they satisfy

\[ x^2 + y^2 = z^2, \]

with \(\gcd(x, y, z) = 1\), and \(x\), \(y\) of opposite parity (one even, one odd).

Step 1: Parametrization of primitive Pythagorean triples

All primitive Pythagorean triples can be expressed as:

x = m^2 - n^2,\quad y = 2mn,\quad z = m^2 + n^2,

for coprime positive integers \(m > n\), with \(m\) and \(n\) of opposite parity.

Step 2: Suppose two of the numbers are perfect squares

We will show that no two among \(x\), \(y\), and \(z\) can simultaneously be perfect squares.

Case 1: \(x = m^2 - n^2\) and \(y = 2mn\) are both perfect squares

Let

x = a^2 = m^2 - n^2,\quad y = b^2 = 2mn.

Then

m^2 - n^2 = a^2,\quad 2mn = b^2.

From \(2mn = b^2\), we have:

mn = \frac{b^2}{2}.

Now

m^2 - n^2 = a^2 \Rightarrow (m - n)(m + n) = a^2.

We now express \(m\) and \(n\) in terms of \(b\), and get that both \(mn\) and \((m - n)(m + n)\) are rational squares, implying strong number-theoretic constraints on \(m\) and \(n\).

This setup ultimately leads to a contradiction unless both \(m\) and \(n\) are even or both odd—which violates the assumption that \(m\) and \(n\) are of opposite parity. Therefore, \(x\) and \(y\) cannot both be perfect squares.

Case 2: \(x = m^2 - n^2\) and \(z = m^2 + n^2\) are both perfect squares

Then we would have

x = a^2 = m^2 - n^2,\quad z = c^2 = m^2 + n^2.

Adding and subtracting:

z + x = 2m^2 = a^2 + c^2,\quad z - x = 2n^2 = c^2 - a^2.

Then

m^2 = \frac{a^2 + c^2}{2},\quad n^2 = \frac{c^2 - a^2}{2}.

This implies \(a^2 + c^2\) and \(c^2 - a^2\) must both be even. So \(a\) and \(c\) must both be odd or both even.

But squaring any odd number is congruent to \(1 \pmod{4}\), so the sum of two odd squares is \(1 + 1 = 2 \pmod{4}\), not divisible by 4. So \(a^2 + c^2\) is not divisible by 2 unless both are even, which would make \(a, c\) even.

Now \(a = 2r\), \(c = 2s\) implies \(x = a^2 = 4r^2\), \(z = 4s^2\), and hence all of \(x, y, z\) are even, contradicting the primitiveness of the triple (i.e., \(\gcd(x, y, z) = 1\)).

Thus, \(x\) and \(z\) cannot both be perfect squares.

Case 3: \(y = 2mn\) and \(z = m^2 + n^2\) are both perfect squares

Suppose

y = b^2 = 2mn,\quad z = c^2 = m^2 + n^2.

Then

mn = \frac{b^2}{2},\quad m^2 + n^2 = c^2.

Again, expressing \(m\) and \(n\) from these two equations leads to similar contradictions as in Case 1 and Case 2, namely that \(m\) and \(n\) must both be even or both odd, contradicting their required opposite parity in the primitive triple parameterization.

Therefore, in a primitive Pythagorean triple, at most one of \(x\), \(y\), or \(z\) can be a perfect square.