Adjacent Pythagorean triples

Question

Is 3, 4, and 5 the only Pythagorean triple where the numbers are adjacent, and if so, why?

Answer 1

Hint: Solve $(n-1)^2+n^2=(n+1)^2$.

Answer 2

Yes, because the equation $n^2+(n+1)^2=(n+2)^2$ is polynomial, thus it can just have that many solutions at most.

Answer 3

Yes. Here's one way to see why: let $(x-1,x,x+1)$ be a Pythagorean triple, then $$(x-1)^2 + x^2 = (x+1)^2$$ This is a quadratic equation, so it can have at most $2$ solutions. $x=0$ and $x=4$ are two solutions, so there are no others. $x=0$ gives the triple $(-1,0,1)$ which is invalid, so the only allowable triple is $x=4$: $(3,4,5)$

Answer 4

As otherwise mentioned solving: $$a^2+(a+1)^2=(a+2)^2\to2a^2+2a+1=a^2+4a+4\to a^2-2a-3=0$$ $$\to(a-3)(a+1)=0\to a=3,a=-1$$ Hence we have $(3,4,5)$, and the invalid $(-1,0,1)$.

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You can also show this with the two-variable pythagorean theorem. $$(p^2-q^2)^2+(2pq)^2=(p^2+q^2)^2$$ What your argument asks is if there are $p,q$ satisfies either of these conditions. $$(p^2+q^2)-(2pq)=1 \space \text{AND}\space (p^2+q^2)-(p^2-q^2)=2$$ $$\text{OR}$$ $$(p^2+q^2)-(2pq)=2 \space \text{AND}\space (p^2+q^2)-(p^2-q^2)=1$$ See if you can show when these work.

Answer 5

Pedestrian way:

$a^2+(a+1)^2=(a+2)^2;$

$(a+2)^2-(a+1)^2=a^2;$

$1 \cdot (2a+3)= a^2;$

$a^2-2a -3=0;$

$(a-3)(a+1)=0;$

$a=3; a=-1;$

Hence: $3,4,5$ is the only triple .