Nonexistence of Pythagorean triple with largest side prime.

Question

The Problem: In the Pythagorean triplets (a,b,c) when a < b then b can't be a prime number.

The Background: While searching the properties of Pythagorean triplets in web I saw quite a few listed, but didn't see the above one which I thought was true, because I had developed a proof.

The Request: As discussed many a times in this site I would request some alternate proofs (or counterexamples) before I share mine for a review.

Answer 1

Hint $\rm\,\ a^2\! + p^2 = c^2\:\Rightarrow\: p^2 = (c\!-\!a)(c\!+\!a).\:$ Unique factorization $\:\Rightarrow \begin{eqnarray}\rm\:c\!-\!a &=&1\\ \rm c\!+\!a &=&\rm p^2\end{eqnarray}\:$ contra $\rm\,a<p$

Answer 2

We use "Euclid's" formula for generating the Pythagorean triples. The job can be done much more elegantly, with much less machinery. Please see Bill Dubuque's answer.

The Pythagorean triples all have shape $k(x^2-y^2)$, $k(2xy)$, $k(x^2+y^2)$. Here $x$ and $y$ are positive integers such that $x \gt y$ (and $x$ and $y$ are relatively prime, and of opposite parity, though these side facts do not matter for the proof). If a leg is to be prime we need $k=1$.

Certainly $2xy$ cannot be prime. And $x^2-y^2=(x-y)(x+y)$ cannot be prime unless $x=y+1$. So from now on we may assume that $x=y+1$.

So $x^2-y^2=(y+1)^2-y^2=2y+1$. But $2xy=2(y-1)(y)=2y^2-2$. We cannot have $2y+1\gt 2y^2-2$. So the leg $2y+1$ must be the smallest. In particular, the larger of the two legs cannot be prime.

Answer 3

We begin with Euclid's formula shown here as $$ A=m^2-k^2\qquad B=2mk \qquad C=m^2+k^2$$ At a glance, we can see that B is even and, in fact, B is always a multiple of $4.\quad$ To prove this, we substitute $\quad m=(2n-1+k)\quad$ and get a formula I developed in $2009$ to generate only the subset of triples where $GCD(A,B,C)$ is an odd square which includes all primitives. $$A=(2n-1+k)^2-k^2\quad B=2(2n-1+k)k \\ C=((2n-1+k)^2+k^2)$$ becomes

\begin{align*} A=(2n-1)^2+ & 2(2n-1)k \\ B= \qquad\quad\quad & 2(2n-1)k+ 2k^2\\ C=(2n-1)^2+ & 2(2n-1)k+ 2k^2\\ \end{align*}

where B is the sum of two even numbers and thus a multiple of $4.\quad$ The formula generates

\begin{array}{c|c|c|c|c|c|} n & k=1 & k=2 & k=3 & k=4 & k=5 \\ \hline Set_1 & 3,4,5 & 5,12,13& 7,24,25& 9,40,41& 11,60,61 \\ \hline Set_2 & 15,8,17 & 21,20,29 &27,36,45 &33,56,65 & 39,80,89 \\ \hline Set_3 & 35,12,37 & 45,28,53 &55,48,73 &65,72,97 & 75,100,125 \\ \hline Set_{4} &63,16,65 &77,36,85 &91,60,109 &105,88,137 &119,120,169 \\ \hline Set_{5} &99,20,101 &117,44,125 &135,72,153 &153,104,185 &171,140,221 \\ \hline Set_{6} &43,24,145 &165,52,173 &187,84,205 &209,120,241 &231,160,281 \\ \hline \end{array}]

By inspection, we can see that $B=4x, x\in\mathbb{N}$ includes all primitives and not just some double or quadruple of primitives. We can also see that half of all triples have $A>B$ and, no matter what, B is still a multiple of $4$. The conclusion is that B can never be prime.

Answer 4

If $k,x,y\in\Bbb Z^+$ and $A=k(x^2-y^2), B=2kxy, C=k(x^2+y^2)$ and $A$ is prime, then, since $A=k(x-y)(x+y)$ is prime with $x+y>1$, we must have $x-y=1.$ So $$\frac B A=\frac {2xy}{x^2-y^2}=\frac {2(y+1)y}{(y+1)^2-y^2}=$$ $$=\frac {2y^2+2y}{2y+1}>\frac {2y^2+y}{2y+1}=y\ge 1.$$