Two Sums of Two Squares

Suppose such natural number $n = a^2 + b^2 = c^2 + d^2$ , where $a, b, c, d$ are some natural numbers.

Then $a^2 - c^2 = d^2 - b^2$

Thus, $(a - c)(a + c) = (d - b)(d + b)$

The difference between the factors on the left hand side is $2c$ and always even, so we can assure that these factors are either odd-odd or even-even. The same applies to the right hand side.

The smallest possible factors are $1\times 3$ . However, $3$ is prime and has no other composite factorization.

Then we move to $2\times 4 = 1\times 8$ . However, with $1$ odd and $8$ even, the difference $7$ can never be even.

Finally, for $3\times 5 = 1\times 15$ , we can set two sets of equations:

$a - c = 3 ; a + c = 5$ $a = 4 ; c = 1$ $d - b = 1 ; d + b = 15$ $d = 8 ; b = 7$

Hence, $4^2 - 1^2 = 8^2 - 7^2$ .

$n = 4^2 + 7^2 = 8^2 + 1^2 = \boxed{65}$

If a number can be written as the sum of two squares in two different ways, it is of the form $(ac + bd)^2 + (bc - ad)^2 = (bc + ad)^2 + (ac - bd)^2 = (a^2 + b^2)(c^2 + d^2).$ We must have $a \not = b$ and $c \not = d$ to make the values really different.

We also want $a$ and $b$ to have opposite parity (even and odd). Otherwise, $a^2 + b^2 = 2 \cdot \left[\left(\frac{a+b}2\right)^2 + \left(\frac{a-b}2\right)^2\right]$ will immediately give us a smaller pair of squares to work with.

The lowest values for which all of this works are $a = 3, b = 2$ and $c = 2, d = 1$ , giving $13 \cdot 5 = \boxed{65} = 8^2 + 1^2 = 7^2 + 4^2.$

For those who like to see this principle at work a little more:

$\begin{array}{cc|cc||rlll} a & b & c & d & \\ \hline 3 & 2 & 2 & 1 & 65 & = 8^2 + 1^2 & = 7^2 + 4^2 & = 13 \cdot 5 \\ 4 & 1 & 2 & 1 & 85 & = 6^2 + 7^2 & = 9^2 + 4^2 & = 17 \cdot 5 \\ 5 & 2 & 2 & 1 & 145 & = 12^2 + 1^2 & = 9^2 + 8^2 & = 29 \cdot 5 \\ 6 & 1 & 2 & 1 & 185 & = 13^2 + 4^2 & = 8^2 + 11^2 & = 37 \cdot 5 \\ 5 & 4 & 2 & 1 & 205 & = 14^2 + 3^2 & = 13^2 + 6^2 & = 41 \cdot 5 \\ 4 & 1 & 3 & 2 & 221 & = 14^2 + 5^2 & = 11^2 + 10^2 & = 17 \cdot 13 \\ \end{array}$