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Let $a,b,c$ form a primitive Pythagorean triple (meaning $(a,b,c) \in \mathbb{Z}^{3}$). Let $a$ be odd. Use a proof by contradiction to show that $(c-b)$ and $(c+b)$ share no common prime factors.

My Thoughts

So I began by assuming that $(c-b)$ and $(c+b)$ shares at least one common factor.

And so, for some prime $p$, $(c-b) = xp$ and $(c+b) = yp$, such that $(x,y) \in \mathbb{Z}^{2}$

So now, I thought maybe if we set this equal to either $a^2$ or $c^2 - b^2$ I could find a contradiction. But in either case, I couldn't come to any conclusion.

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    $\begingroup$ At math.stackexchange, it is generally required that you demonstrate at least some of your own effort in solving the problem. This is done so that you yourself can see and understand where you have gotten stuck to try to avoid this in the future. $\endgroup$
    – cws
    Feb 17 '18 at 0:07
  • $\begingroup$ Good job adding some detail. $\endgroup$
    – Joffan
    Feb 17 '18 at 0:17
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You were on the right track.

From $c-b = xp\;$and $c+b = yp$, you get $$a^2 = c^2 - b^2 = (c+b)(c-b) = xyp^2$$ hence $p^2{\,\mid\,}a^2$, so $p{\,\mid\,}a$.

Since $a$ is odd, and $p{\,\mid\,}a$, it follows that $p$ is odd.

From $p{\,\mid\,}(c+b)$ and $p{\,\mid\,}(c-b)$, we get $p{\,\mid\,}\bigl((c+b)-(c-b)\bigr)$ and $p{\,\mid\,}\bigl((c+b)+(c-b)\bigr)$.

Thus, $p{\,\mid\,}2b$ and $p{\,\mid\,}2c$, but then, since $p$ is odd, it follows that $p{\,\mid\,}b$ and $p{\,\mid\,}c$.

Then $p$ is a common factor of $a,b,c$, contrary to the assumption that the triple $(a,b,c)$ is primitive.

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  • $\begingroup$ What if p=2? Then you can't deduce p|b or p|c because b and c could be odd. $\endgroup$ Feb 17 '18 at 0:21
  • $\begingroup$ Thanks -- now fixed. Oddness of $a$ was key, else using $a=4,b=3,c=5$ would yield a counterexample. $\endgroup$
    – quasi
    Feb 17 '18 at 0:44
  • $\begingroup$ Are you quasi from sci.math ?? $\endgroup$
    – mick
    Feb 17 '18 at 1:00
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    $\begingroup$ @mick: Yes.${}$ $\endgroup$
    – quasi
    Feb 17 '18 at 1:11
  • $\begingroup$ Omg , you might know my mentor then. $\endgroup$
    – mick
    Feb 17 '18 at 1:14
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So now, $xp\cdot yp = (c+b)(c-b)=c^2-b^2 = a^2$ which is odd (since $a$ is given as odd).

Thus $(c+b)$ and $(c-b)$ are both odd also. as are $x,y,p$. And $c = (xp+yp)/2 = p(x+y)/2$ is divisible by $p$, as is $a$, and thus $p$ must divide $b$ also, which contradicts the primitive nature of the triple.

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And so, for some prime p, (c−b)=xp and (c+b)=yp, such that (x,y)∈Z2

So now, I thought maybe if we set this equal to either a2 or c2−b2 I could find a contradiction. But in either case, I couldn't come to any conclusion

If your first idea was to let $p$ be prime so that $x*p = c-b$ and $y*p = c =b$, then

Your first trick should be to note is $a^2 + b^2 = c^2 \implies a^2 = c^2 - b^2 = (c+b)(c-b)=xyp^2$ so $p|a$

The second trick is note that $p|(c-b)$ and $c|c+b$ means $p|(c-b) + (c+b) = 2c$ and $p|(c+b)-(c-b) = 2b$. So if $p\ne 2$ then $p$ divides all of $a,b,$ and $c$.

And then the last idea is to note that the problem did state that $a$ was odd (so $p$ can't equal $2$), and that $a,b,c,$ were primitive (so no such prime can exist).

(Although it could if $a$ were even and $b$ and $c$ were odd and $p=2$$\ ^*$. Or, of course, if $a,b,c$ weren't primitive and therefore any common divisor would have to also be a common divisor of $c-b$ and $c+b$.)

Those tricks will become more and more apparent and familiar with experience.

$\ ^*$ If $a$ is even then $2$is the only factor $c+b$ and $c-b$ will have in common. $b$ and $c$ will both have to be odd (if one were even so would the other) and so $c+b$ and $c-b$ are both even. One of $c+b$ or $c-b$ is divisible by $4$ but the other will not be.

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Another way:

Using https://en.m.wikipedia.org/wiki/Pythagorean_triple#Generating_a_triple,

let $a=m^2-n^2,b=2mn, c=m^2+n^2$ where $(m,n)=1$ and $m,n$ have opposite parity

Now if prime $p$ divides both $c\pm b=(m\pm n)^2,$

$p$ must be odd and must divide $m\pm n$

and hence will divide $m+n\pm(m-n)$

and hence $2(m,n)$

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Since $a$ is odd, $b$ is even . Because if $a=2a'+1$ is odd and if $b=2b'+1$ is also odd then $c^2=(2a'+1)^2+(2b'+1)^2$ has a remainder of $2$ modulo $4,$ which is impossible for a square.

So $c^2$ is odd, so $c$ is odd . So $c+b$ and $c-b$ are both odd.

Let $d$ be a common divisor of $c+b$ and $c-b$. Note that $d$ must be odd.

Then $d$ divides $(c+b)+(c-b)=2c$ and $d$ divides $(c+b)-(c-b)=2b.$ But $d$ is odd, so $d$ must divide $c$ and $b .$

Since $d$ divides $c+b$ and $c-b$ it divides the product $(c+b)(c-b)=a^2.$ But if $d$ is prime and $d$ divides $a^2$ then $d$ divides $a.$ This would make $d$ a common prime divisor of $a,b,c,$ which would make $(a,b,c)$ non-primitive.

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I developed a formula that generates only and all triples where $(C-B)=(2n-1)^2,n\in\mathbb{N}$. This subset includes all primitives and none of the trivials, the doubles and the square multiples of primitives that Euclid's formula generates.

$$A=(2n-1)^2+2(2n-1)k\quad =2(2n-1)k+2k^2\quad C=(2n-1)^2+2(2n-1)k+2k^2$$

This generates only primitives except when $k$ is multiple of any factor of $(2n-1)$. For this discussion, $(2n-1)$ does not divied $k$ because we require a primitive triple.

We can see by definition that $$\frac{C+B}{C-B}=\frac{\big(2(2n-1)k+2k^2\big)+\big(2n-1)^2+4(2n-1)k+4k^2\big)}{(2n-1)^2} =\frac{(2n-1)^2+4(2n-1)k+4k^2}{(2n-1)^2}$$ The denominator shares factors with the first two terms of the numerator but not with the third term of the numerator so $(C-B)$ and $(C+B)$ have no common factors.

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