Numbers that are the sum of 2 distinct nonzero squares in exactly 1 way I need to emulate this sequence for a program: http://oeis.org/A025302
Stuff that I've taken into account:


*

*After finding the prime divisors of a number. I take any divisor as p and apply the following rule:


p is an odd prime divisor of n, then either p appears to an even power in n, or p≡1(mod4)


*

*Also I only consider sums of 2 distinct nonzero squares


The sequence in the described link only goes as far as 229. Until here my simulator using these 2 rules work, but I'm not so sure after that since I get these numbers ( which I'm not sure that they belong in my sequence ) when I loop until n <= 1000..
[0] 325 int
[1] 425 int
[2] 650 int
[3] 725 int
[4] 845 int
[5] 850 int
[6] 925 int

So my question is:
Do these 7 digits belong originally to the http://oeis.org/A025302 sequence, or have I gotten wrong my algorithm, if so, can you point the missing rule ?
Any question, feel free to comment
 A: Copying from http://mathworld.wolfram.com/SumofSquaresFunction.html: 

To find in how many ways a positive integer $n>1$ can be expressed as a sum of $k=2$ squares ignoring order and signs, factor it as
  $$n=2^{a_0}p_1^{2a_1}...p_r^{2a_r}q_1^{b_1}...q_s^{b_s}$$
  where the $p_i$ are primes of the form $4k+3$ and the $q_i$ are primes of the form $4k+1$. If $n$ does not have such a representation with integer $a_i$ because one or more of the powers of $p_i$ is odd, then there are no representations. Otherwise, define
   $$B=(b_1+1)(b_2+1)...(b_r+1)$$
  The number of representations of $n$ as the sum of two squares ignoring order and signs is then given by:  
$$
r'_2(n)=
\left\{
  \begin{array}{lll}
   0 & \textrm{if any }  a_i \textrm{ is a half integer},  \\ 
\frac{1}{2}B & \textrm{if all }  a_i \textrm{ are integers and } B \textrm{ is even}, \\ 
\frac{1}{2}\big(B-(-1)^{a_0}\big) & \textrm{if all }  a_i \textrm{ are integers and } B \textrm{ is odd}
  \end{array}
\right.
$$

which implies that the missing rule in your algorithm (which btw is just the Sum of two squares theorem), is the following:
In order to have $r'_2(n)=1$, there can only be three cases: 


*

*Either: $n$ has a single $4k+1$ prime factor with multiplicity $b=1$, hence  $B=b+1=1+1=2$ and $r'_2(n)=\frac{1}{2}B=1$
($n=5$ provides an example for that case),

*or: $n$ has a single $4k+1$ prime factor with multiplicity $b=2$ and the multiplicity (in the prime factorization of $n$) of $2$ is $a_0$:even, hence  $B=b+1=2+1=3$ and $r'_2(n)=\frac{1}{2}(B-(-1)^{a_0})=\frac{1}{2}(3-1)=1$
($n=100=2^2\cdot 5^2$ provides an example for that case), 

*or: there are no $4k+1$ prime factors in the prime factorization of $n$ and the multiplicity (in the prime factorization of $n$) of $2$ is $a_0$:odd, hence in that case, all $b_j$ are zero which gives $B=1$ thus $r'_2(n)=\frac{1}{2}(B-(-1)^{a_0})=\frac{1}{2}(1+1)=1$.
However, this last case should be excluded from your considerations since you are asking for numbers $n$ which can be written in a unique way as a sum of two distinct squares while this last case leads to numbers which are twice a square ($n=18=2\cdot 3^2=3^2+3^2$ for example). 


You might also be interested in Theorem 4.4, p. 10 from these notes and as for some code the function PowersRepresentations[n, 2, 2] could also be of some use to your purposes (however note that this function also allows zeros). 
