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If it is, is it trivial?

                                          **Theorem**

There does not exist a pythagorean triple $a^2 + b^2 = c^2$ $\{a,b,c \in \Bbb N\}$ where $b \ge a$ and $a|b$

                                          **PROOF**

Suppose $\exists$ a constant k {$k | k \ge 1, k \in \Bbb N$} and $k \cdot a$ forms a pythagorean triple with a.

I.e $a^2 + (k \cdot a)^2 = c^2$

$a^2 + k^2 \cdot a^2 =c^2 $

$a^2(k^2 + 1) = c^2$

$c = \sqrt{a^2(k^2 + 1)}$

$c = a \sqrt{k^2 + 1} $

$ \exists $ $ k: \sqrt{k^2 + 1} \in \Bbb Q$

However $\forall$ $ a \in \Bbb N$, $ a \ge 1$ $ \sqrt{a^2 + 1} \notin \Bbb Q$*

Where $ \Bbb Q$ is the set of rational numbers.

*: The proof of this theorem is left as an exercise to the reader.

Tobi Alafin
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  • I hope b|a reads a is a factor of b? – Tobi Alafin Nov 17 '16 at 14:16
  • It is usually read $b$ divides $a$. Your proof is correct otherwise (once you change the theorem's hypothesis) – Vincent Nov 17 '16 at 14:19
  • So I should change it to a|b ? – Tobi Alafin Nov 17 '16 at 14:24
  • What I meant in english is this: $ b \ge a $ and a is a factor of b. – Tobi Alafin Nov 17 '16 at 14:29
  • Then it should be $a|b$ indeed. – Sil Nov 17 '16 at 14:32
  • The proof shows only that $\sqrt{k^2+1}$ cannot be $c/a$ (rational). How did you deduce that it would have to be natural number? For example you can have $1 = 2 \cdot \frac{1}{2}$ – Sil Nov 17 '16 at 14:39
  • I said $k: k \ge 1$ I don't think $\exists$ $k \in \Bbb N$ $k: (\sqrt{k^2 +1} ) \in \Bbb N$ I have an informal proof, but I'm not sure how rigorous it is, which is why I left it as an exercise to the reader. I am not following you. – Tobi Alafin Nov 17 '16 at 14:44
  • Two comments: If $a|b$ then $b \ge a$. There have been a number of proofs here showing that the square root of every non-square integer is irrational. – marty cohen Nov 17 '16 at 14:55
  • My point is that you have shown that $c = a \sqrt{k^2 + 1} $, and then somehow from that you concluded that $\sqrt{k^2+1}$ must be $\in \Bbb N$, which does not follow. It just has to be rational, and as marty mentions, there are proofs for that not being possible. – Sil Nov 17 '16 at 14:57
  • Oh. I get what you're saying now. I'll edit my question accordingly. – Tobi Alafin Nov 17 '16 at 15:33
  • @marty cohen, we have $\sqrt{K^2 + 1}$ won't you have to prove that this is first a non-square integer. I.e $\forall$ $a \in \Bbb N$ $a^2 + 1$ is \not\eq b^2 – Tobi Alafin Nov 17 '16 at 15:56
  • $\neq b^2$ $ b: b \in \Bbb N$ – Tobi Alafin Nov 17 '16 at 16:05
  • No. $k^2+1$ is a non-square integer because the next square is $(k+1)^2=k^2+2k+1$. – marty cohen Nov 17 '16 at 19:35

1 Answers1

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Let $$\quad A=m^2-n^2\quad B=2mn\quad \text{then} \quad \frac{B}{A}=\frac{2mn}{(m+n)(m-n)}$$ By inspection, we can see that $\frac{B}{A}\in\mathbb{Q}$ but is never an integer or vice versa.

poetasis
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