As indicated in Nate's answer, an explanation comes from a combination of known facts of Galois theory and algebraic number theory. Let $K\subseteq \Bbb{C}$ be the splitting field of $f(x)$ over $\Bbb{Q}$, and let $G=Gal(K/\Bbb{Q})$ be the Galois group. If $X$ is the set of complex zeros of $f(x)$, we know that $G$ faithfully permutes the $n=\deg f(x)$ numbers in $X$. A key fact is that unless the prime $p$ is a factor of the discriminant, the factoring of $f(x)$ modulo $p$ is governed by how a certain automorphism $\sigma_p\in G$ (called the Frobenius automorphism at $p$) acts on the set $X$. More precisely,
the degrees of the factors of $f(x)$ modulo $p$ are exactly the lengths of the cycles of $\sigma_p$, when the latter is viewed as a permutation of $X$. See e.g. Qiaochu Yuan's old post for more.
My main observation is the following.
The degrees of the factors of $f(x)$ modulo $p$ are all equal to each other (with possible exceptions for the factors of the discriminant, see the caveat in the end) if and only if the Galois group $G$
of the polynomial has order $n:=\deg f(x)$.
Furthermore, this is the case exactly when the action of $G$ on $X$ is regular. In other words, if $\alpha$ and $\beta$ are any two elements of $X$, there is a unique element $\sigma\in G$ such that
$\sigma(\alpha)=\beta$. In yet other words, the action on $X$ is equivalent to the Cayley action of $G$ on itself.
We need the action to be regular, for otherwise the action on $X$ will have non-trivial point stabilizers (by the orbit-stabilizer theorem). If $\sigma\in G,\sigma\neq 1_G,$ stabilizes one of the roots, and is also the Frobenius automorphism for some $p$, we immediately get variation in the lengths of the cycles of $\sigma$, and hence also in the degrees of the factors. Conversely, if the action of $G$ is regular, then each automorphism $\sigma\in G$ has cycles of length $ord(\sigma)$ only. Observe that Chebotarev's theorem guarantees that each and every element of $G$ (up to conjugacy) occurs as the Frobenius automorphism for infinitely many primes $p$.
If $G$ is abelian, then this automatically happens. For if $K$ is the splitting field and $\alpha\in K$ is a zero of $f(x)$, then $\Bbb{Q}(\alpha)$ is the splitting fiedl $K$, hence $[K:\Bbb{Q}]=\deg f$, and the main observation bites.
Both of the OP's examples fall under this umbrella. The Galois group of
$x^4+1$ is isomorphic to $\Bbb{Z}_8^*$ as the splitting field is the eight cyclotomic field. Similarly, the Galois group of $x^4-10x^2+1$ is the Klein four, as $\Bbb{Q}(\sqrt2,\sqrt3)$ is the splitting field.
I initially suspected abelianity to be the key, but couldn't prove it, and arrived at the regularity requirement instead. Just as well, see the next paragraph :-).
The Galois group $G$ need not be abelian. For example the minimal polynomial of $\root3\of2+e^{2\pi i/3}$ is
$$f(x)= 9 + 9 x + 3 x^3 + 6 x^4 + 3 x^5 + x^6.$$
Obviously its Galois group is $\simeq S_3$, acting regularly on the set of zeros. Then $f$ splits into a product of either linear, quadratic or cubic factors modulo $p$ all depending on whether the Frobenius automorphism modulo $p$ is the identity, a 2-cycle or a 3-cycle. The cases occur with (asymptotic) frequencies $1/6$, $3/6$ and $2/6$ as there is a single identity element, three $2$-cycles and two $3$-cycles in the group $S_3$.
The construction in the previous paragraph works for every finite group $G$ that can be realized as the Galois group $G=Gal(K/\Bbb{Q})$. This is conjectured (but unknown) to be possible for every finite group. Anyway, when $K=\Bbb{Q}(\rho)$, we can use the minimal polynomial of
$\rho$ as $f(x)$, when the factorization of $f(x)$ will obey this law.
On the other hand, if we add the requirement that the polynomial $f(x)$ remains irreducible modulo some prime $p$, this forces the presence of an $n$-cycle in the group $G$, immediately implying that $G$ must be cyclic.
A nagging concern is the primes that occur as factors of the discriminant. The theory above doesn't cover them. For those primes $p$ the polynomial $f(x)$ has repeated factors (which is why the theory breaks down), and I don't see why the degrees of the irreducible modular factors would still need to be equal. However, that is still the case with the OP's quartics as well as my sextic (IIRC).
But consider as a further example, this irreducible degree 120 polynomial
$$f(x)=x^{120}+33750 x^{114}+1116000 x^{112}+3093750 x^{110}+1113560865 x^{108}+26921565000
x^{106}+604881000000 x^{104}+41167612108500 x^{102}+1431025760469805
x^{100}+14099233308750000 x^{98}+840755432509572240 x^{96}+29314339633598287500
x^{94}+838066825456291786500 x^{92}+17055343730004169787750
x^{90}+447099790210227436061835 x^{88}+16392596904188900682547500
x^{86}+524905496075736354421632150 x^{84}+7746834021558485383973580750
x^{82}+227400983306935610033504910210 x^{80}+5438203936818046114019096970000
x^{78}+161737233618167288338366207390005 x^{76}+3079140294664647133063741473941250
x^{74}+71589845771247618831377143880007030 x^{72}+1703874280827990134916438519652329750
x^{70}+37765960575062006944570272099577441065
x^{68}+671553629574346789742980518018455265750
x^{66}+13425663013791814923118775212295423224575
x^{64}+298316535549382310619276177470147839567500
x^{62}+6263322665754902028679045968192687021946417
x^{60}+113241137056665611046577975852478978853633750
x^{58}+2020482344097260796483156681326789886674641275
x^{56}+33607134753772181350504921906082658777563400000
x^{54}+550090827539075781453680855277159942042963975675
x^{52}+9645703546363285715097504428728612658988657633750
x^{50}+167589376219621416661551144351153197328368374710525
x^{48}+2591935957193938557000893428986050845007482057327500
x^{46}+37023543980552299462242594649218811608185860378407900
x^{44}+467204073589927694863531699904773814089803461948544000
x^{42}+5479652136885621906960743248094128630901237665341582235
x^{40}+60126566311045736532762730255787195693544124509832792500
x^{38}+622160800556257347660482211684147114157184154675733033935
x^{36}+5910280916004942423618805942747770827509420868385152115000
x^{34}+52106693211491045864419719017582977817447744056381867707300
x^{32}+432630947000379472605449339454013771610886258014994363460750
x^{30}+3410774365180782497074812580214046939106788562491870798046950
x^{28}+24817712899719418559089094578783591417923276718955280439927500
x^{26}+166698014114867874997691254686109786765856603463497453170639800
x^{24}+1049202472505126328549462956322404941429926499583868890532339750
x^{22}+6200097454408267204060687491470661741711907284917745068688392736
x^{20}+33063248055138491333939800416229050428492869440711652594093046250
x^{18}+160736778936639982917274226049744035868667050374165352429934665765
x^{16}+713664496954265807293435573649973706894373908524000311681369366250
x^{14}+2807053376943637105788469622562984789953566579091832894551656335585
x^{12}+9405202706949914238281707642961512160794793530858137713058729194750
x^{10}+29874334500240270890684107985135817331403471254102147813332994672740
x^8+70498858873024053142911958048672442662399336143969727445260372215750
x^6+152058512419218423143371634526878068936948757145171468466368924947025
x^4+244691421102204282324955998149483634966444221492246674851974602623750
x^2+191416245283609068525605381407966277298503971734124842355884163512481$$
I (or Mathematica, if truth is to be told) cooked up for this answer should also split into equal degree polynomials modulo $p$. The possible orders of elements of $S_5$ are $1,2,3,4,5$ and $6$, so no higher degree polynomials will appear in the factorizations.
Indeed,
- Modulo $2$ this polynomial is a product of quintic factors, both repeated $12$ times.
- Modulo $3$ it mostly factors in to quartics, all repeated four times, but there is also a quadratic factor, repeated four times.
- Modulo $7$ is factors into cubics, some simple, some repeated twice.
- Modulo $11$ it factors into distinct quintics.
- Modulo $13$ it factors into distinct sextics.
- But modulo $5$ it is mostly a product of fifth powers of quintics, but also of fifth powers of linear factors.
So the nice theory fails in the sense that there may be a finite number of exceptional primes where there is variation among the modular factors.
