Fermat's Two Squares Theorem/Uniqueness Lemma/Proof 1

Lemma for Fermat's Two Squares Theorem

Let $p$ be a prime number.

Suppose there were an expression:

$p = a^2 + b^2$

where $a$ and $b$ are positive integers.

Then that expression would be unique except for the order of the two summands.

Proof

Suppose:

$p = a^2 + b^2 = c^2 + d^2$

where $a, b, c, d \in \Z_{>0}$.

We have that:

\(\ds \paren {a c + b d} \paren {a d + b c}\)

\(=\)

\(\ds \paren {a^2 + b^2} c d + \paren {c^2 + d^2} a b\)

multiplying out and gathering terms

\(\ds \)

\(=\)

\(\ds p \paren {a b + c d}\)

as $p = a^2 + b^2 = c^2 + d^2$

$p$ is prime by hypothesis.

From Euclid's Lemma for Prime Divisors:

$p \divides \paren {a c + b d}$

or:

$p \divides \paren {a d + b c}$

Without loss of generality, suppose $p \divides \paren {a c + b d}$.

By Absolute Value of Integer is not less than Divisors: Corollary:

$(1): \quad a c + b d \ge p$

We have:

\(\ds p^2\)

\(=\)

\(\ds \paren {a^2 + b^2} \paren {c^2 + d^2}\)

by hypothesis

\(\ds \)

\(=\)

\(\ds \paren {a c + b d}^2 + \paren {a d - b c}^2\)

Brahmagupta-Fibonacci Identity

\(\ds \leadsto \ \ \)

\(\ds p^2\)

\(\ge\)

\(\ds p^2 + \paren {a d - b c}^2\)

from $(1)$: $p \le a c + b d$ and so $p^2 \le \paren {a c + b d}^2$

\(\ds \leadsto \ \ \)

\(\ds 0\)

\(\ge\)

\(\ds \paren {a d - b c}^2\)

subtracting $p^2$ from both sides

\(\ds \leadsto \ \ \)

\(\ds a d - b c\)

\(=\)

\(\ds 0\)

Square of Real Number is Non-Negative

\(\text {(2)}: \quad\)

\(\ds \leadsto \ \ \)

\(\ds \dfrac c a\)

\(=\)

\(\ds \dfrac d b\)

dividing both sides by $a b$ and rearranging

\(\ds \leadsto \ \ \)

\(\ds \dfrac {c^2} {a^2}\)

\(=\)

\(\ds \dfrac {d^2} {b^2}\)

squaring both sides

\(\ds \)

\(=\)

\(\ds \dfrac {c^2 + d^2} {a^2 + b^2}\)

Mediant is Between: Corollary $2$

\(\ds \)

\(=\)

\(\ds \dfrac p p\)

by hypothesis

\(\ds \)

\(=\)

\(\ds 1\)

Hence:

$c = a$

and:

$b = d$

$\blacksquare$