Natural number which can be expressed as sum of two perfect squares in two different ways? Ramanujan's number is $1729$ which is the least natural number which can be expressed as the sum of two perfect cubes in two different ways. But can we find a number which can be expressed as the sum of two perfect squares in two different ways. One example I got is $50$ which is $49+1$ and $25+25$. But here second pair contains same numbers. Does any one have other examples ? 

Any Pythagorean triple $(A^2+B^2=C^2)$ provides a candidate where the $C^2$ can have $2$-or-more combinations of $A^2$ and $B^2$ that add up to it. These can be found by testing natural numbers of the form $(4n+1)$ with a range of $m$ values defined as shown below to see which, if any yield integers. We begin with Euclid's formula and solve the $C$-function for $n$ in terms of $C$ and $m$.
$$A=m^2-n^2\quad B=2mn\quad C=m^2+n^2$$
$$C=m^2+n^2\implies n=\sqrt{C-m^2}\\
\text{where}\qquad \bigg\lfloor\frac{ 1+\sqrt{2C-1}}{2}\bigg\rfloor \le m \le \big\lfloor\sqrt{C-1}\space\big\rfloor$$
The lower limit ensures $m>n$ and the upper limit ensures $n\in\mathbb{N}$.
$$C=65\implies\\
 \bigg\lfloor\frac{ 1+\sqrt{130-1}}{2}\bigg\rfloor=6 \le m \le \big\lfloor\sqrt{65-1}\big\rfloor=8\\
\text{ and we find} \quad m\in\{7,8\}\Rightarrow n\in\{4,1\}\\$$
$$F(7,4)=(33,56,65)\quad  F(8,1)=(63,16,65) $$
Here, we have $\space 33^2+56^2=65^2\space$ and $\space 63^2+16^2=65^2$
There infinitely many of these pairs of triples; the next that meet the criteria are
$$(13^2,84^2,85^2),\qquad  (77^2,36^2,85^2)\\ (75^2,100^2,125^2),\qquad  (117^2,44^2,125^2)\\(17^2,144^2,145^2),\qquad  (143^2,24^2,145^2)\\ ...$$
$\textbf{Edit:}$ If $\space n \space$ is the number of distinct prime fators of $\space C,\space$ there are $2^{n-1}$ primitive triples with that same
$C$-value. This means e.g. that for  $1105=5\times13\times17,\space$ there are $2^{3-1}=4$ primitives with $C=1105.$
