Pythagorean triples and primes.

Question

Determine whether there are any right-angled triangles with integer lengths such that the lengths of both of the sides adjacent to the right angle are primes.

This was the question I was asked by my teacher. My first answer is no, which I still believe is correct, but then she asked me "why?" and I got stumped. My first thought was that Pythagorean triples can be generalized as: $$2k,\quad k^2-1,\quad k^2+1$$ Is this generalization correct? Because, if so, $2k$, for $k > 1$, cannot be a prime as it is divisible by $2$. Even if $k= 1$, there are no Pythagorean integer triples that have $2$.

Also, $k^2-1$ cannot be a prime, for $k > 2$, because it can be factored like this: $(k+1)(k-1)$ Even if $\quad k = 2,\quad 2k = 4,\space k^2-1=3, \space k^2+1 = 5.\quad$ Thus, the two sides adjacent to the right angle would be 3 and 4, one of which (4) is not prime.

Only $k^2+1$ could be prime, proving that either a or b can be prime, but both can't.

Is this proof correct? If the generalization is wrong, the proof is also flawed. Thanks.

Answer 1

If $a=2$ and $b\ge 2$ is prime, we need an integer $c$ with $c^2=b^2+2^2$. As $(b+1)^2=b^2+2b+1>b^2+4$, we’d need $b^2<c^2<(b+1)^2$, which is impossible.

Hence we must have odd sides $a=2r+1$ and $b=2s+1$, so $$\tag1c^2=(2r+1)^2+(2s+1)^2 =4(r^2+r+s^2+s)+2,$$ a number that is even, but not a multiple of $4$. If $c$ is odd, then $c^2$ is odd; and if $c$ is even, $c^2$ is a multiple of $4$. Neither of these options is compatible with $(1)$, hence no solution exists.

Answer 2

The most referenced and accepted way of generating Pythagorean triples is by using Euclid's formula $\quad (A=m^2-k^2\quad B=2mk\quad C=m^2+k^2) \quad m,k\in\mathbb{N}\quad $ as it produces a subset including all "primitive" triples, $\space GCD(A,B,C)=1,\space$ the only ones considered interesting.$\space\space$ Imprimitives contain no primes. We can reason that side-B cannot be prime unless $\space B=2\space$ and that this cannot be true because the smallest triple is $\space (3,4,5).$

We can more strongly reason that $\space B\not\in\mathbb{P}\space$ because, in all Pythagorean triples, $\quad A=2x+1,\space B=4y,\space C=4z+1\quad x,y,z\in\mathbb{N}.\quad $ To visualize this, consider a table of sets of triples that includes all primitives generated by a variation of Euclid's formula:

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

\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 \end{array}

Side-A can be prime in $\space Set_1\space$ such as when $A\in\{3,5,7,11,13,\cdots\space$ but $\space n>1\implies A\not\in\mathbb{P}\space\space $because $\space (2n-1)^2+2(2n-1)k=(2 n - 1) (2 k + 2 n - 1) \longrightarrow\space$ $\space n>1\implies (2n-1)\in\{3,5,7,9,\cdots\} \implies$ side-A is composite.

$\text{Side-} B\not\in\mathbb{P}\space$ because $ 2(2n-1)k +2k^2=2 k (k + 2 n - 1),\space 2 k\implies B\text{ even and } (k + 2 n - 1)\ge 2\implies B\ge4. \quad$ Side=C can be prime such as when $\space C\in\{5,13,17,29,37,41,53,61,\cdots\}.\quad $

Side-A and side-C can both be prime but side-B is always composite.

$\therefore\quad$ The sides adjacent to the right angle cannot both be primes.