Are there finitely many Pythagorean triples whose smallest two numbers differ by 1?

Has it been shown whether there is a finite or infinite number of Pythagorean triples whose smallest two numbers differ by 1?

In either case I’d appreciate a link to the proof.

Edit: thank you all for your answers. Would it also be possible to determine for which triples the smallest number is prime?

There is an infinite number of such pythagorean triples. Any primitive pythagorean triple $(a,b,c)$ with $a^2+b^2=c^2$ (are we are looking for primitive triples since we want $a$ and $b$ consecutive) is of the form: $$ a=p^2-q^2,\qquad b=2pq,\qquad c=p^2+q^2 $$ with $p$ and $q$ coprime and not both odd. So we are looking for integer solutions of: $$ p^2-2pq-q^2 =\pm 1,$$ or: $$ (p-q)^2 - 2q^2 = \pm 1.$$ However we know that the Pell equation $A^2-nB^2=1$ has an infinite number of integer solutions $(A,B)$ for every $n$ that is not a square, hence we can find "consecutive" pythagorean triples from the solutions of $$ A^2 - 2B^2 = 1,\tag{1}$$ for istance. $(A,B)=(3,2)$ is the minimal solution of $(1)$, giving $(p,q)=(5,2)$, hence the triple $(20,21,29)$. The next solution can be found by expanding: $$ (3+2\sqrt{2})^2 = 17+12\sqrt{12},$$ hence $(p,q)=(29,12)$ gives the triple $(696,697,985)$ and so on.

In general, we can see that all the solutions depends on the convergents of the continued fraction of $\sqrt{2}$, i.e. on the Pell sequence: $$ (p,q) = (P_n,P_{n+1}),$$ from which:

$$ (a_n,b_n) = (2P_nP_{n+1},P_{n+1}^2-P_{n}^2) = (2P_n P_{n+1},2P_nP_{n+1}+(-1)^n),$$

where:

$$P_n = \frac{1}{2\sqrt{2}}\left((1+\sqrt{2})^n-(1-\sqrt{2})^n\right).$$

Since neither $2P_n P_{n+1}$ or $P_{n+1}^2-P_{n}^2$ may be primes if $n> 1$, the only "consecutive" pythagorean triple with the smallest element being a prime is $(3,4,5)$.

Let's suppose $(x,x+1,y)$ is a Pythagorean triple. Then:

$x^2+(x+1)^2 = y^2$

$2x^2+2x+1 = y^2$

$4x^2+4x+2 = 2y^2$

$(2x+1)^2 + 1 = 2y^2$

$(2x+1)^2 - 2y^2 = -1$

This is now a negative Pell's Equation for which there are infinitely many solutions. See this question for details.

Cheap version: we get triples $(x,y,z)$ such that $y=x+1$ and $x^2 + y^2 = z^2,$ with $$ (3 ,4,5 ) $$ $$ (20 ,21,29 ) $$ $$ ( 119,120, 169 ) $$ $$ (696 ,697, 985 ) $$ $$ (4059 , 4060,5741 ) $$ $$ (23660 ,23661,33461 ) $$ $$ (137903 ,137904,195025 ) $$

That is, $(x_n, y_n,z_n)$ such that $$ x_{n+2} = 6 x_{n+1} - x_n +2, $$ but $$ y_{n+2} = 6 y_{n+1} - y_n -2, $$ and $$ z_{n+2} = 6 z_{n+1} - z_n . $$

If you put them the way I had them at first, odd first and even second, you get a $4 (-1)^n$ addend that I thought was a bit too much work.