What are the four positive rational numbers whose fourth powers add up to the integer $34996$?

The question asks to find rationals $x_i$ such that,

$$x_1^4+x_2^4+x_3^4+x_4^4 = N$$

for 14 integers $N <100000$, namely,

$N = 34996, 44516, 46516, 50756, 54836, 55796, 58916, 63076, 63556, 75236, 78036, 85956, 97876, 98596$

missing from the OP's initial search. Solutions were found for almost all (except the last). The first is,

$$ 34996=\left(\frac{48005}{5593}\right)^4+ \left(\frac{71165}{5593}\right)^4+ \left(\frac{42575}{5593}\right)^4+ \left(\frac{2731}{5593}\right)^4 $$

This is equivalent to the integer equality

$$ 34996\times 5593^4 = 48005^4+71165^4+42575^4+2731^4 $$

which can be checked directly. Also, as

$$34996\times y_0^4=y_1^4+y_2^4+y_3^4+y_4^4$$

then $\gcd(y_0,y_1,y_2,y_3,y_4)=1.$ The rest of the solutions are,

$$ \begin{aligned} 44516 &= \left(\frac{15649}{5363}\right)^4 +\left(\frac{27705}{5363}\right)^4\\ &\quad+ \left(\frac{77535}{5363}\right)^4 +\left(\frac{13775}{5363}\right)^4 \end{aligned} $$

$$ \begin{aligned} 46516 &= \left(\frac{66449}{4641}\right)^4 +\left(\frac{27535}{4641}\right)^4\\ &\quad+ \left(\frac{35035}{4641}\right)^4 +\left(\frac{6705}{4641}\right)^4 \end{aligned} $$

$$ \begin{aligned} 50756 &= \left(\frac{61989}{4159}\right)^4 +\left(\frac{19345}{4159}\right)^4\\ &\quad+ \left(\frac{23005}{4159}\right)^4 +\left(\frac{1455}{4159}\right)^4 \end{aligned} $$

$$ \begin{aligned} 54836 ={}&\left(\frac{46891}{6657}\right)^4 +\left(\frac{50675}{6657}\right)^4 \\ &+\left(\frac{70915}{6657}\right)^4 +\left(\frac{91785}{6657}\right)^4 \end{aligned} $$

$$ \begin{aligned} 55796 ={}&\left(\frac{74707}{7971}\right)^4 +\left(\frac{6755}{7971}\right)^4 \\ &+\left(\frac{70535}{7971}\right)^4 +\left(\frac{114075}{7971}\right)^4 \end{aligned} $$

$$ \begin{aligned} 58916 ={}&\left(\frac{33319}{5871}\right)^4 +\left(\frac{34805}{5871}\right)^4 \\ &+\left(\frac{63845}{5871}\right)^4 +\left(\frac{84375}{5871}\right)^4 \end{aligned} $$

$$ \begin{aligned} 63076 ={}&\left(\frac{28757}{4699}\right)^4 +\left(\frac{85}{4699}\right)^4 \\ &+\left(\frac{52715}{4699}\right)^4 +\left(\frac{68755}{4699}\right)^4 \end{aligned} $$

$$ \begin{aligned} 63556 ={}&\left(\frac{24649}{6069}\right)^4 +\left(\frac{55955}{6069}\right)^4 \\ &+\left(\frac{62905}{6069}\right)^4 +\left(\frac{88155}{6069}\right)^4 \end{aligned} $$

$$ \begin{aligned} 75236 ={}&\left(\frac{64961}{4221}\right)^4 +\left(\frac{32545}{4221}\right)^4 \\ &+\left(\frac{35835}{4221}\right)^4 +\left(\frac{42635}{4221}\right)^4 \end{aligned} $$

$$ \begin{aligned} 78036 ={}&\left(\frac{43931}{4559}\right)^4 +\left(\frac{3895}{4559}\right)^4 \\ &+\left(\frac{28575}{4559}\right)^4 +\left(\frac{73585}{4559}\right)^4 \end{aligned} $$

$$ \begin{aligned} 85956 ={}&\left(\frac{29309}{4379}\right)^4 +\left(\frac{28615}{4379}\right)^4 \\ &+\left(\frac{29745}{4379}\right)^4 +\left(\frac{73645}{4379}\right)^4 \end{aligned} $$

$$ \begin{aligned} 97876 ={}&\left(\frac{48857}{5857}\right)^4 +\left(\frac{39485}{5857}\right)^4 \\ &+\left(\frac{63785}{5857}\right)^4 +\left(\frac{97535}{5857}\right)^4 \end{aligned} $$

$$ \begin{aligned} 98596 ={}&\left(\frac{20387}{10977}\right)^4 +\left(\frac{27985}{10977}\right)^4 \\ &+\left(\frac{33185}{10977}\right)^4 +\left(\frac{194445}{10977}\right)^4 \end{aligned} $$

Clearing denominators will yield integer equalities similar to the first one. All 14 solved now.

Method

Here is a short description of the search method used by my code.

We search for primitive integer identities of the form $$ Nq^4=a^4+b^4+c^4+d^4, \qquad \gcd(a,b,c,d,q)=1, $$ which give rational solutions after division by $q^4$. For the values of $N$ considered here we have $$ N\equiv 4\pmod {16}, \qquad N\equiv 1\pmod 5. $$ The congruence modulo $16$ implies that $q$ is odd: otherwise the left hand side is $0\bmod 16$, forcing all four numerators to be even, contrary to primitivity. Once $q$ is odd, the left hand side is $4\bmod 16$, so all four of $a,b,c,d$ must be odd. Similarly, the congruence modulo $5$ implies $5\nmid q$, and then exactly one of $a,b,c,d$ is non-divisible by $5$. Thus, after reordering, the code only has to search equations of the form $$ Nq^4=a^4+(5b)^4+(5c)^4+(5e)^4, \qquad 5\nmid a . $$

For each fixed denominator $q$, the congruence modulo $625$ gives the strong restriction $$ a^4\equiv Nq^4\pmod {625}, $$ because the three terms $(5b)^4,(5c)^4,(5e)^4$ vanish modulo $625$. So the code first computes the admissible residue classes for $a\bmod 625$, rather than trying all possible values of $a$.

The main search is then a meet-in-the-middle computation. Write $$ a^4+(5e)^4 = Nq^4-(5b)^4-(5c)^4. $$ For the fixed value of $q$, the code stores all possible left-hand-side values $a^4+(5e)^4$ in an exact hash table, with $a$ restricted to the allowed residue classes modulo $625$. It then runs through the pairs $(b,c)$ and checks whether $$ Nq^4-(5b)^4-(5c)^4 $$ appears in that table. This changes the search from a four-variable search to two quadratic searches, which is the main speedup.

All reported identities are then checked by the exact equality $$ a^4+(5b)^4+(5c)^4+(5e)^4=Nq^4. $$ The additional Bloom and residue filters in the implementation are only speed filters: they avoid many hash-table lookups, but every reported solution is still verified by the exact integer identity above.