Is a reducible coprime binomial always a difference of squares? | Mathematics | Page 1

last cloak Jul 6, 2025, 3:03 AM

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Is a reducible coprime binomial always a difference of squares, or something similar?
This is for Q[x,y...]

This is what I've done so far (! marks assumptions/givens, ? marks goal/question):

! (a+b)(c+d) = J+K : gcf(J,K)=1
! J and K are monomials

(a+b)(c+d) = ac+ad+bc+bd
! a+b ≠ 0, a ≠ 0, b ≠ 0
! c+d ≠ 0, c ≠ 0, b ≠ 0

ac + ad = 0 -> a(c+d) = 0, contridiction (WLOG {kx,ky})
-> ac+bd = 0 WLOG
[ No zero divisors | SEE integral domain ]

ac = -bd
p (prime) exists because Q[x,y...] such that p | ac 
[ SEE Unique Factorization Domain ]
p | ac -> p | bd
p | a or c and p | b or d
WLOG: {p | a and p | b} or {p | a and p | c}

{p | a and p | b}
-> A,B exists: a = pA, b = pB
-> (a+b) = (pA + pB) = p(A+B)
-> J + K = (a+b)(c+d) = p(A+B)(c+d)
-> p | J + K, which are terms, therefore
-> p | gcf(J,K) -> gcf(J,K) ≠ 1 contridiction

{p | a and p | c}
-> A,C exists: a = pA, c = pC
J + K = (a+b)(c+d) = (pA+b)(pC+d)

? (c+d) = ±(a-b)

reef coyoteBOT Jul 6, 2025, 3:03 AM

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last cloak Jul 6, 2025, 3:17 AM

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wait, a ~ c have a special relationship from ac+bd = 0 asymmetric to a ~ d that declares the past WLOG ( {p | a or p | c} ) incomplete

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oh that step was really wrong,

WLOG: {p | a and p | b} or {p | a and p | c}```

i should've written `d` instead of `c`. that was that problem.

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okay, now with corrections it has become:

{p | a and p | d}
-> A,D exists: a = pA, d = pD
J + K = (a+b)(c+d) = (pA+b)(c+pD)

(pA+b)(c+pD) = p²AD+pAc+pDb+bc = ad+bc
-> ad = p²AD+pAc+pDb
-> p²AD = p²AD+pAc+pDb
-> 0 = pAc+pDb
-> 0 = Ac+Db = ac+bd

and it seems promising

last cloak Jul 6, 2025, 3:54 AM

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Beginning to repeat!

Ac = -Db
q (prime) exists because Q[x,y...] such that q | Ac 
[ SEE Unique Factorization Domain ]
q | Ac -> q | Db
q | A or c and q | D or b
WLOG: {q | A and q | b}, {q | A and q | D}, or {q | b and q | c}

{q | A and q | b}
-> q | A and A | a -> q | a
-> q | a and b
-> q | (a+b) -> q | J+K -> q | gcf(J,K), contradiction (same as earlier)

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{q | A and q | D}
-> A₂,D₂ exist: A = qA₂, D = qD₂
J + K = (a+b)(c+d) = (pA+b)(c+pD) = (pqA₂+b)(c+pqD₂) // a = pA = pqA₂
(pqA₂+b)(c+pqD₂) = p²q²A₂D₂+pqA₂c+pqD₂b+bc = ad+bc
-> ad = p²q²A₂D₂+pqA₂c+pqD₂b
-> p²q²A₂D₂ = p²q²A₂D₂+pqA₂c+pqD₂b
-> 0 = pqA₂c+pqD₂b
-> 0 = A₂c+D₂b = Ac+Db = ac+bd```
If this case is forced now, it will continue to be forced, leading to infinite reduction via a prime. That would be a big lead.

last cloak Jul 7, 2025, 3:22 AM

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I have completed it:

...
3. {q | b and q | c}
-> B,C exist: b = qB, c = qC
J + K = (a+b)(c+d) = (a+qB)(qC+d)
(a+qB)(qC+d) = q²BC+qCa+qBd+ad = ad+bc
-> bc = q²BC+qCa+qBd
-> q²BC = q²BC+qCa+qBd
-> 0 = qCa+qBd
-> 0 = Ca+Bd = ac+bd
--------------------------------------------------------------
The following repeats, untill units A*, B*, C*, D* are found such that
a = PA*, b = QB*, c = QC*, d = PD*
where for example, p | P and q | P or Q
and A*C* = -B*D*

(a+b)(c+d) = (PA*+QB*)(QC*+PD*) = J + K
(PA*+QB*)(QC*+PD*) = PQA*C* + PQB*D* + P²A*D* + Q²B*C* = ad+bc
-> PQA*C* + PQB*D* + P²A*D* + Q²B*C* = P²A*D* + Q²B*C*
-> PQA*C* + PQB*D* = 0
-> PQ(-B*D*) + PQ(B*D*) = 0

so what has been learnt?
(a+b)(c+d) = (PA*+QB*)(QC*+PD*)
-> a and d are like-terms
-> b and c are like-terms
-> J+K = P²A*D*+Q²B*C*

in Z[x,y...], there are finite units: {-1,1}
-> (a+b)(c+d) = (PA*+QB*)(PC*+QD*)
A*C* = -B*D*
same signs     = + -> different signs 
different signs = - -> same signs
same signs     = + -> different signs
different signs = - -> same signs
-> one pair has same signs, other has different signs
WLOG: A* = C* and B* = -D*
WLOG: let the sign of A* be +; - -> + by P -> -P
WLOG: let the sign of B* be +; - -> + by Q -> -Q
(a+b)(c+d) = (P+Q)(P-Q) = J + K
P²-Q² = J + K
Q.E.D. (note: necessary to be in Z[x,y...])

? (c+d) = ±(a-b)

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full:

📎 binomialalwaysDOS.txt

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.close

reef coyoteBOT Jul 7, 2025, 3:23 AM

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Solved

Post marked as solved by @last cloak.

Use .unsolved if this was a mistake.

last cloak Jul 9, 2025, 3:38 AM

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i just realized this works for Q[x,y...] as well, as difference of squares

(PA*+QB*)(QC*+PD*) and PQA*C* = -PQB*D* (which immediately leads to A*C* = -B*D*)
WLOG, let the units A* and B* be 1
-> (P+Q)(QC*+PD*) and C* = -D*
-> (P+Q)(QC*+P(-C*))
-> (P+Q)(QC*-PC*)
-> (Q+P)(Q-P)C*

meaning that it is always in the form of difference of squares, scaled by a unit (for all UFDs I believe)

last cloak Jul 15, 2025, 2:39 AM

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found counterexample: x³-1 = (x-1)(x²+x+1)

this happened because my proof assumed that polynomials cannot have factors with greater amount of terms, which is demonstrated to be incorrect.

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.unsolved

reef coyoteBOT Jul 15, 2025, 2:39 AM

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Unsolved

Post marked as unsolved by @last cloak.

Use .solved to mark as solved.

ocean furnace Jul 15, 2025, 4:56 PM

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I'm interested in your problem, but i can't quite get it, mind explaining it a tiny bit?

last cloak Jul 19, 2025, 2:08 AM

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i just realized that i marked this as unsolved when i found it was wrong, and the question "Is a reducible coprime binomial always a difference of squares?" was answered with "No; ex: x³-1 = (x+1)(x²+x+1)"

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.solved

reef coyoteBOT Jul 19, 2025, 2:08 AM

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Solved

Post marked as solved by @last cloak.

Use .unsolved if this was a mistake.

last cloak Jul 19, 2025, 2:09 AM

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ocean furnace I'm interested in your problem, but i can't quite get it, mind explaining it a t...

basically i was thinking of all the ways you can factor a binomial

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and i knew some were prime, and some were a constant factor of a prime, and some were difference of squares, and I didn't know about what the rest consisted of

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(i also didn't know of the "difference/sum of cubes" patterns before this; if i knew those, it would of immediently answered my question with examples like x³-1)

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and i conflated the fact that factors of polynomials have equal or lower degree with an untrue statement that factors of polynomials have equal or lower amount of terms, leading me to assume binomials either have constant factors (which are trivial to find) or other binomial factors

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and i knew the difference of squares [ x²-y² = (x+y)(x-y) ] was one method to find binomial factors of binomials, so i wanted to know if it found all of them when used recursively [ ex: x⁴-y⁴ = (x²+y²)(x²-y²) = (x²+y²)(x+y)(x-y) ]

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so
"Is a reducible coprime binomial always a difference of squares, or something similar?"
meant

reducible -> this method either finds the factors, or if it can't be applied, this proved the original binomial was already prime
coprime -> to specify that any constant factor has been removed ( a+b where a and b are coprime; primitive might've been a better word?)
to be a difference of squares -> a+b = x²-y² of some x and y, where either it is a = x² or b.

tldr: "Is the difference of squares method the only method you need to factor a binomial?"

#Is a reducible coprime binomial always a difference of squares?