# Elementary Proof Relating to Pythagorean Triples

Please could I get some feedback on the correctness and overall quality of the following proof, which is meant to show that all Pythagorean triples contain unique integers.

To Prove: all Pythagorean triples contain unique integers.

First we note that in a right-angled triangle, if $$c$$ is the hypotenuse and $$a,b$$ are the remaining sides, $$c > a$$ and $$c > b$$ by triangle inequality. Now it remains to prove that

$$\forall (a,b,c) \in \mathbb{N}$$, where $$a^2+b^2=c^2$$, $$a \ne b$$

Assume that $$a = b$$. Then we have $$a^2 + a^2 = c^2$$

Thus, $$c = \sqrt{2a^2} = a\sqrt{2}$$

Since an integer multiple of an irrational number remains irrational, $$c \notin \mathbb{N}$$, which is a contradiction of our original assumption.

Therefore $$a \ne b$$

Since $$a \ne b$$ and $$c > a$$ and $$c > b$$,

$$a,b,c$$ are distinct integers and theorem is proven.

$$\square$$

• It's a small detail, but maybe specify that $a,b,c \in \mathbb{Z}^+$. Sometimes, $0$ is considered to be a natural number, and this trivial case doesn't work.
– D.B.
May 5, 2019 at 17:15
• I don't like the phrase "contains unique integers". You should say "contains 3 distinct integers". When something is unique it means it's the only one of its kind, e.g. there is a unique $x\in \Bbb R$ such that $x^3=8.$ In mathematical English, when we say $a,b,c$ are distinct , we mean $a\ne b\ne c\ne a$......+1. May 6, 2019 at 5:19

I think your proof seems to work fine. However, maybe it’s worth mentioning when you state that ‘an integer multiple of an irrational number remains irrational’, you are excluding the case where that integer is zero. Obvious, but worth mentioning for rigour. Perhaps state at the beginning that $$a,b,c$$ are positive integers.

The only other thing I think you could make clearer is how you know that $$c>a$$ and $$c>b$$. The triangle inequality actually says that $$a + b \geq c$$ (or $$a+b>c$$ if you exclude degenerate triangles), so perhaps you should give a few more steps to show how you arrive at your conclusion.

I would say that you're proof is fine!

There's only a claim that I would change.

$$c>a$$ and $$c>b\:$$ by the triangle inequality.

However, the triangle inequality simply states that the inequality $$a+b>c$$ is true for all its permutations...

$$c=\sqrt{a^2+b^2}>\sqrt{a^2+0}=a\iff c>a$$

You may also want to be more rigorous and prove the claim

Since an integer [nonzero-]multiple of an irrational number remains irrational...

You can, for instance, use contradiction again.

Let $$x\in\Bbb N, y\in\Bbb R\setminus \Bbb Q$$ and suppose that for some $$z\in\Bbb Q$$ such that $$z=\frac pq$$ (recall the definition of rational) $$x\cdot y=z\iff y=\frac{p}{qx}$$ which is a contradiction, since irrational numbers cannot be expressed as a fraction between to integers...

We are given$$\quad A=m^2-n^2\quad B=2mn\quad C=m^2+n^2.$$ and we wish to prove that that $$A,B,C$$ are unique integers.

If $$A=B$$ then$$\quad(m^2-n^2)^2+(m^2-n^2)^2=C^2\Rightarrow C=(m^2-n^2)\sqrt{2}$$ so it is not Pythagorean triple where all sides are integers.

If $$B=A$$ then$$\quad (2mn)^2+(2mn)^2=8m^2n^2=C\Rightarrow C=2mn\sqrt{2}$$ so it is also not Pythagorean triple where all sides are integers.

If $$A=C$$ then$$\quad m^2-n^2=m^2+n^2\Rightarrow 0=2n^2$$ and any zero value of $$n$$ will produce a trivial triple where $$B=0$$ such as $$(1,0,1)$$.

If $$B=C$$ then $$\quad 2mn=m^2+n^2\Rightarrow m^2-2mn+n^2=(m-n)^2=0$$. This can only be true if $$m=n$$ and that will always produce a trivial triple where $$A=0$$ such as $$(0,1,1)$$

None of these conditions produce a Pythagorean triple. Therefore, $$A,B,C$$ are always distinct integers.