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 ?

Note that $a^2 + b^2 = c^2 + d^2$ is equivalent to $a^2 - c^2 = d^2 - b^2$, i.e. $(a-c)(a+c) = (d-b)(d+b)$. If we factor any odd number $m$ as $m = uv$, where $u$ and $v$ are both odd and $u < v$, we can write this as $m = (a-c)(a+c)$ where $a = (u+v)/2$ and $c = (v-u)/2$. So any odd number with more than one factorization of this type gives an example.

Thus from $m = 15 = 1 \cdot 15 = 3 \cdot 5$, we get $8^2 - 7^2 = 4^2 - 1^2$, or $1^2 + 8^2 = 4^2 + 7^2$.

From $m = 21 = 1 \cdot 21 = 3 \cdot 7$ we get $11^2 - 10^2 = 5^2 - 2^2$, or $2^2 + 11^2 = 5^2 + 10^2$.

$$ 65 = 64 + 1 = 49 + 16 $$

This will work for any number that's the product of two primes each of which is congruent to $1$ mod $4$. For more than two ways multiply more than two such primes.

The following example easily generalizes:

$$\begin{align} 5&=(2+i)(2-i)=4+1\\ 13&=(3+2i)(3-2i)=9+4\\ 5\cdot13&=((2+i)(3+2i))((2-i)(3-2i))=(4+7i)(4-7i)=16+49\\ &=((2+i)(3-2i))((2-i)(3+2i))=(8-i)(8+i)=64+1 \end{align}$$

$2465$ can be expressed as the sum of two squares in four different ways:- $8^2 + 49^2$, $16^2 + 47^2$, $23^2 + 44^2$ and $28^2 + 41^2$.