By giving you a Right triangle with Hypotenuse of 35 units, what are the possible integer values for the sides of the right triangle?
It's easy to know the integer values when we know one of the right triangle sides by factoring $a^2+b^2=c^2 \Leftrightarrow a^2=(b+c)(b-c)$ but when $c$ is given we can't factor $a^2+b^2$ in real numbers.
On
We have a theorem that if prime $p=4k+3$ and $p|a^2+b^2$ then $p|a$ and $p|b$ so in our case $p=7$. So we have $a=7x$ and $b=7y$, so we have $x^2+y^2 = 25$ and for this we have well know solutions 3 and 4.
On
We know that if $a$, $b$ and $c$ are sides-lengths of the right-angled triangle, where $c$ is a length of the hypotenuse and $a$ is even, then there are naturals, $m$, $n$ and $d$, where $m>n$, $gcd(m,n)=1$ and $m$ and $n$ have the opposite parity, for which $a=2mnd$, $b=(m^2-n^2)d$ and $c=(m^2+n^2)d$.
Thus, $(m^2+n^2)d=35$ and the rest is smooth.
we have $$a^2+b^2=35^2$$ by $AM-GM$ we get $$\frac{a^2+b^2}{2}\geq ab$$ so we get $$ab\le \frac{35^2}{2}$$ if $$b\le a$$ we get $$b^2\le ab\le \frac{35^2}{2}$$