Theorem
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Status: proved exact reformulation
Date: 2026-08-15
Depends on: ES-TWO-TARGET-SIGNED-BOX-EQUIVALENCE.md, FAB-UNBOUNDED-DIVISOR-RATIO-CERTIFICATE.md
Claim boundary: this is an equivalent coordinate form of the exact prime Type-I/Type-II signed-box theorem. It does not prove universal target existence.
1. From signed exponents to divisors of C^2
Let
be prime and let
Put
The signed divisor box is
For each signed exponent vector define
Then
and
runs through every positive divisor of C^2 exactly once as the exponent vector varies.
Moreover
Therefore
This is a bijective reparametrization of the signed box.
2. Type-I target
By inversion symmetry, the Type-I target may be written as
instead of -p^{-1}.
Since
the equation
is equivalent to
Because C is a unit modulo k, this becomes
or
Thus:
Exact Type-I divisor-square criterion
This recovers the fixed-k divisor-square target from the earlier strong FAB lane, now as the exact standard Type-I coordinate.
3. Type-II target
The Type-II target is
Hence
is equivalent to
Therefore:
Exact Type-II divisor-square criterion
This is the divisor-square counterpart that was missing from the one-target fixed-k formulation.
4. Exact two-target theorem
Combining the two cases with ES-TWO-TARGET-SIGNED-BOX-EQUIVALENCE.md gives:
Theorem — prime ES as two divisor classes inside Div(C^2)
For prime p≡1 mod4, Erdős--Straus holds for p if and only if there exists
such that, with
there is a divisor d|C^2 satisfying at least one of
or
Equivalently,
The first class is Type I; the second is Type II.
5. Complement involution
The divisor set has the natural involution
Under the signed-box map
this is exactly inversion:
Type I
The two orientations -p and -p^{-1} correspond to a complementary divisor pair.
If
then
which is the divisor coordinate of the inverse Type-I orientation.
Type II
The Type-II residue is self-inverse. If
then
Thus the Type-II target class is fixed by divisor complement.
This self-reciprocity is a useful structural distinction between the two solution types.
6. Type-II square-pair form
Suppose
Put
Then also
Since de=C^2 is a square, d and e have the same squarefree kernel. Write
with s squarefree. Then
Now
Because every prime factor of sa divides C and gcd(C,k)=1, one has
Therefore
So Type II may equivalently be viewed as a complementary divisor-square pair whose square roots add to a multiple of the shift.
This recovers the familiar normalized factor-pair geometry from a different direction.
7. Strategic consequence
The exact prime problem can be stated without signed exponents:
Choose an admissible shift
k. The square divisor lattice ofC=(p+k)/4must hit one of two residue classes modulok: the fixed Type-I class-4^{-1}or the moving Type-II class-C.
The two classes have different involution behavior:
- Type I is a complementary pair under
d -> C^2/d; - Type II is self-complementary.
This form may be better suited to divisor-distribution, reciprocity, and lattice arguments than the original signed-box notation while remaining exactly equivalent to it.