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Let $s, t \in \mathbb{Z}^+$, be both odd, $s > t \geq 1$, $\gcd(s, t) = 1$. If we set

$$a = st,\quad b = \frac{1}{2}(s^2 - t^2),\quad c = \frac{1}{2}(s^2 + t^2)$$

then $(a, b, c)$ is primitive pythagorean triple.

I tried to prove it but I am not sure of my method. Could you please check my solution if there exists something senseless.

What I tried:

Suppose there exists a $p$ which is prime number, such that $p\ |\ b$ and $p\ |\ c$.

$$p\ |\ b = \frac{1}{2}(s^2 - t^2) \Rightarrow p\ |\ 2b = s^2 - t^2 \Rightarrow p\ |\ s^2\ \textrm{and}\ p\ |\ t^2 \Rightarrow p\ |\ s\ \textrm{and}\ p\ |\ t\\ p\ |\ c = \frac{1}{2}(s^2 + t^2) \Rightarrow p\ |\ 2c = s^2 + t^2 \Rightarrow p\ |\ s^2\ \textrm{and}\ p\ |\ t^2 \Rightarrow p\ |\ s\ \textrm{and}\ p\ |\ t$$

But we know that $\gcd(s, t) = 1$ and if $p\ |\ s$ and $p\ |\ t$, then this is contradiction.

Is this true?

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$p|(s+t) \Rightarrow p|s\: \text{and} p|t\quad $ is never true. –  adam W Oct 7 '12 at 21:39
    
@adamW so how should I show that p|s and p|t? –  Xentius Oct 7 '12 at 21:45
    
since p is prime and divides $a=st$ then can only say for sure it only divides $s$ or $t$. Besides, gcd(s,t)=1 is given, nothing divides both. –  adam W Oct 7 '12 at 21:49
    
@adamW sorry I do not understand. Do you start with the assumption that p is prime and p|a, then go on, or do you find that p|a after making some other assumptions in the beginning? Could you please write the proof from the beginning to the end? Sorry for my ignorance. –  Xentius Oct 7 '12 at 21:53
    
@adam. A minor nit, but in your first comment, that $p\;|\;s+t\Rightarrow p\;|\;s\text{ and }p\;|\;t$ is never true, I'd strike the "never", since there are some cases when the expression is indeed true. You're right, of course, that the implication, considered as a single logical expression, is false. –  Rick Decker Oct 11 '12 at 0:50
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3 Answers 3

up vote 0 down vote accepted

After the point that $p\ |\ \frac{1}{2}(s^2-t^2)$ and $p\ |\ \frac{1}{2}(s^2+t^2)$ assume that, $\frac{1}{2}(s^2-t^2) = pk$ and $\frac{1}{2}(s^2+t^2) = pl$. Then, if you add two equations: $s^2 = p(k+l) \Rightarrow p\ |\ s^2 \Rightarrow p\ |\ s$.

Do the same thing for $t$ and show that $p$ also divides $t$. Then if $p\ |\ s$ and $p\ |\ t$ it should also divide $a$ but in this case $(a, b, c)$ is not a PPT. (Contradiction.)

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If $p|b$ and $p|c$ then it must be true that $p|a$ and therefore the triple is not primitive.

The equations are built to satisfy the pythagorean triple, $a^2 + b^2 = c^2$ (that I assume is the correct placement of the variables), and the gcd restrictions are there to keep scale multiples of the equation from consideration, those scale multiples would be the ones considered not primitive.

So other than checking the equations work, all that is left is to make sure the equation is primitive, which is to say not a multiple. I will leave the algebra for checking the equations, but as far as divisibility goes, the information I have given should give you good food for thought.

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BTW, p does not divide both b and c. A proof of that could be given. Or the contradiction that you reach could prove your starting assumption that p|b and p|c is wrong. –  adam W Oct 7 '12 at 22:03
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I think it is easier to prove instead that gcd $(a,b)=1$.

If $p|a$ then $p|st$ thus $p|s$ or $p|t$.

Since $p|b$ then $p|s^2-t^2=(s-t)(s+t)$. Thus $p|s \pm t$.

Now it is very easy to prove that ($p|s$ or $p|t$) and $p|s \pm t$ implies that $p$ divides both $s$ and $t$.

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