Fermat's Two Squares Theorem/Uniqueness Lemma/Proof 3
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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
For $p = 2$ the only possibility is:
- $p = 1^2 + 1^2$
Assume $p \ge 3$.
Let $a \ge b$ and $c \ge d$ satisfy:
- $p = a^2 + b^2 = c^2 + d^2$
As $2 \nmid p$, we have:
- $(1): \quad a > b$ and $c > d$
Observe:
| \(\ds \paren {a c + b d} \paren {a d + b c}\) | \(=\) | \(\ds \paren {a^2 + b^2} c d + \paren {c^2 + d^2} a b\) | ||||||||||||
| \(\ds \) | \(=\) | \(\ds p \paren {a b + c d}\) |
On the other hand:
| \(\ds p^2\) | \(=\) | \(\ds \paren {a^2 + b^2} \paren {c^2 + d^2}\) | by hypothesis | |||||||||||
| \(\ds \) | \(=\) | \(\ds \paren {a c - b d}^2 + \paren {a d + b c}^2\) | Brahmagupta-Fibonacci Identity: Corollary | |||||||||||
| \(\ds \) | \(>\) | \(\ds \paren {a d + b c}^2\) | as $a c - b d > 0$ by $(1)$ and Real Number Ordering is Compatible with Multiplication | |||||||||||
| \(\ds \leadsto \ \ \) | \(\ds p\) | \(\nmid\) | \(\ds a d + b c\) |
Hence:
- $p \divides a c + b d$
In particular:
- $p \le a c + b d$
Thus:
| \(\ds p^2\) | \(=\) | \(\ds \paren {a^2 + b^2} \paren {c^2 + d^2}\) | by hypothesis | |||||||||||
| \(\text {(2)}: \quad\) | \(\ds \) | \(=\) | \(\ds \paren {a c + b d}^2 + \paren {a d - b c}^2\) | Brahmagupta-Fibonacci Identity | ||||||||||
| \(\ds \) | \(\ge\) | \(\ds p^2 + \paren {a d - b c}^2\) | ||||||||||||
| \(\ds \leadsto \ \ \) | \(\ds a d - b c\) | \(=\) | \(\ds 0\) |
Inserting this back into $(2)$, we obtain:
- $p^2 = \paren {a c + b d}^2$
That is:
- $p = a c + b d$
Hence:
| \(\ds \paren {a - c}^2 + \paren {b - d}^2\) | \(=\) | \(\ds a^2 + b^2 + c^2 + d^2 - 2 p\) | ||||||||||||
| \(\ds \) | \(=\) | \(\ds 2 p - 2p\) | by hypothesis | |||||||||||
| \(\ds \) | \(=\) | \(\ds 0\) |
Thus $a = c$ and $b = d$.
$\blacksquare$