Shortest other leg in a right triangle with one leg of length 2001

[lfdahl]

The shorter leg of an integer-sided right triangle has length 2001. How short
can the other leg be?

 

[Albert1]

The shorter leg of an integer-sided right triangle has length 2001. How short can the other leg be?

my solution:

Spoiler

$2001=667\times 3=3\times 23\times 29$
so the other leg =$667\times 4=2668$

 

[lfdahl]

Thankyou Albert for the correct answer - and for your tireless participation!(Clapping)

Suggested solution:

Spoiler

Let $a,b,c$ be the sides of the triangle. Thus $2001 = a < b < c$. Set $c=b+m$. Then $(b+m)^2 = b^2+2001^2$ or $m(2b+m) = 2001^2$. So $m$ is a divisor of $2001^2=3^2\cdot 667^2$ and since $b= c-m$ is to be shortest ($> 2001$), $m = 667$ (the next largest divisor is $3 \cdot 667 = 2001$, which makes $b=0$) should be considered. Then $667(2b+667) = 9\cdot667^2$ gives $b=2668$ and $c=2668 + 667=3335$. One checks, that $2001^2+2668^2 = 3335^2$.
Comment: This triangle is the $(3,4,5)$ triangle since $(2001,2668,3335) = 667(3,4,5)$. But recognizing this, does not prove, that $2668$ is the shortest possible side larger than $2001$.