Handy Rules For Determining Whether an Integer is the Sum of Two Fourth Powers I would like to develop a set of rules  for determining whether an integer is the sum of two fourth powers.  So far, I have managed to deduce five such rules.  These are:
Rule Number One:  If an integer is the sum of two fourth powers, then all of its prime factors will be congruent to 1 (mod 8) . Equivalently, all of its prime factors will be congruent to 1  or  9  (mod 16).
Rule Number Two: Integers which are the sum of two fourth powers cannot be congruent to 3  or  4 (mod 5).
Rule Number Three: Integers which are the sum of two fourth powers cannot be congruent to  7 ,  8   or  11 (mod 13).
Rule Number Four: Integers which are the sum of two fourth powers cannot be congruent to  6, 7 , 10  or  11 (mod 17).
Rule Number Five: Integers which are the sum of two fourth powers cannot be congruent to  4  ,  5 , 6 , 9 ,  13 , 22  or  28 (mod 29).
Ideally ,  I would like to have between  10  and 15  such rules, which would help me in determining whether an integer  is the sum of two  fourth powers. If anyone can add to this list  then please  do so .  If you know any property or characteristic  of integers which are sums of two fourth powers  then please  tell me about it.
Clearly  there are infinitely many such rules which can be derived , for  congruence's modulo a prime , but for practical reasons ,  10  to 15  will be sufficient.  If you can , add to my "List of Rules".  Thank you in advance.    
 A: Rule one is violated by $2^4+4^4=272=2^4\cdot 17$ and $3^4+6^4=1377=3^4\cdot 17$.  You can have any prime factor as long as it appears a multiple of four times.  You can extend the list of rules two through five as long as you wish, but sums of two fourth powers are quite rare.  Even if you satisfy those rules the chance any large number is a sum of two fourth powers is quite low.
A: Something that I think rules out many numbers
$a^4 \equiv 0,1 \mod 16 \implies a^4 + b^4 \equiv 0,1,2\mod 16$
A: Note that Ross Millikan's notion of rarity can be made explicit: since there are $O(n^{1/4})$ fourth powers $\lt n$, there are at most $O(n^{1/4}\times n^{1/4}) = O(n^{1/2})$ numbers $\lt n$ which are the sum of two fourth powers. But by the Chinese Remainder theorem, any finite number of congruence relations can only eliminate all but $\alpha n$ of the numbers $\lt n$, for some $\alpha\gt 0$; since $\alpha n\gg O(n^{1/2})$, these heuristics don't reduce the haystack you're searching for your needle in by any (asymptotically) appreciable amount.
