Bound on the difference-of-squares representation of a natural number

Question

1. Let $$m^2-n^2=r,$$ where $\gcd (m,n) =1,\;m>n>0,\;$ $m$ and $n$ are of opposite parity, and $r$ is an odd positive integer.

   Then, according to my study materials, determining all the $(m,n)$ solutions involves systematically trying every $n$ such that $1\leq n<\frac r2.$

   (For example: given $m^2-n^2=21,$ the only solutions $(5,2)$ and $(11,10)$ are determined by trying $n=1,2,3,...,10.)$

   But why the upper bound $\frac r2$ for $n\,?$

2. On a related note, for $m^2+n^2=r$ instead, I am able to see that $1\leq n<\sqrt \frac r2.$

   (Please correct me, though, if I'm wrong.)

P.S. This Question is related (though not a duplicate).

Answer 1

If $(m+n)(m-n)=d$ and we take $m>n>0$, we can see that the largest difference between the two factors occurs when $m+n=d$, and $m-n=1$.

Taking the difference we get $2n=d-1$ so that the greatest possible value of $n$ is $\frac {d-1}2$.

Answer 2

The differences between squares increase as the squares themselves increase. If $n,\ m > 10$ then the smallest difference possible is $$12^2 - 11^2 = (12 - 11)(12 + 11) = 23$$ which is already larger than your answer.

By the way, these are not Pythagorean triples.

Answer 3

$r=m^2-n^2\ge(n+1)^2-n^2=2n+1$, so $n\le(r-1)/2$.