We have searched how a powered number could systematically be shared into a sum of 2 coprime numbers. From binomial, we have studied different ways of grouping terms together so that
∀
n
,
(
x
+
y
)
n
=
u
n
+
v
n
,
u
n
∧
v
n
=
1
{\displaystyle \forall n,~(x+y)^{n}=u_{n}+v_{n},~u_{n}\wedge v_{n}=1}
. With odd
n
{\displaystyle n}
and
(
x
,
y
)
{\displaystyle (x,y)}
coprime of opposite parity, we have found out two possibilities. They involve the same
f
n
(
x
,
y
)
{\displaystyle f_{n}(x,y)}
functions that we must now introduce.
Definition
Let us define
f
n
(
x
,
y
)
{\displaystyle f_{n}(x,y)}
functions as
n
=
2
m
+
1
,
f
n
(
x
,
y
)
=
∑
k
=
0
m
(
n
2
k
)
x
m
−
k
y
k
{\displaystyle \quad n=2m+1,~f_{n}(x,y)=\sum _{k=0}^{m}{n \choose 2k}x^{m-k}y^{k}}
Example
f
3
(
x
,
y
)
=
x
+
3
y
f
5
(
x
,
y
)
=
x
2
+
10
x
y
+
5
y
2
f
7
(
x
,
y
)
=
x
3
+
21
x
2
y
+
35
x
y
2
+
7
y
3
f
9
(
x
,
y
)
=
x
4
+
36
x
3
y
+
126
x
2
y
2
+
84
x
y
3
+
9
y
4
.
.
.
{\displaystyle \quad {\begin{array}{lll}f_{3}(x,y)=x+3y\\f_{5}(x,y)=x^{2}+10xy+5y^{2}\\f_{7}(x,y)=x^{3}+21x^{2}y+35xy^{2}+7y^{3}\\f_{9}(x,y)=x^{4}+36x^{3}y+126x^{2}y^{2}+84xy^{3}+9y^{4}\\...\end{array}}}
Algebraic properties
edit
Propositions
(
x
−
y
)
n
=
x
f
n
(
x
2
,
y
2
)
−
y
f
n
(
y
2
,
x
2
)
(
1
)
(
x
−
y
)
n
=
x
f
n
(
x
,
y
)
2
−
y
f
n
(
y
,
x
)
2
(
2
)
{\displaystyle \quad {\begin{array}{lll}(x-y)^{n}&=xf_{n}(x^{2},y^{2})&-&yf_{n}(y^{2},x^{2})&\quad (1)\\(x-y)^{n}&=xf_{n}(x,y)^{2}&-&yf_{n}(y,x)^{2}&\quad (2)\end{array}}}
Proof
Binomial theorem gives:
∀
n
,
(
x
+
y
)
n
=
∑
k
=
0
n
(
n
k
)
x
n
−
k
y
k
{\displaystyle \forall n,~(x+y)^{n}=\sum _{k=0}^{n}{n \choose k}x^{n-k}y^{k}}
Here
n
{\displaystyle n}
is odd. So (1) is simply obtained by grouping together the odd power of
x
{\displaystyle x}
and
y
{\displaystyle y}
(2) is a consequence of (1).
Indeed it gives
(
x
+
y
)
n
=
x
f
n
(
x
2
,
y
2
)
+
y
f
n
(
y
2
,
x
2
)
{\displaystyle (x+y)^{n}=xf_{n}(x^{2},y^{2})+yf_{n}(y^{2},x^{2})}
Thus by multiplying:
(
x
−
y
)
n
(
x
+
y
)
n
=
x
2
f
n
(
x
2
,
y
2
)
2
+
y
2
f
n
(
y
2
,
x
2
)
2
{\displaystyle (x-y)^{n}(x+y)^{n}=x^{2}f_{n}(x^{2},y^{2})^{2}+y^{2}f_{n}(y^{2},x^{2})^{2}}
And finally
(
x
2
−
y
2
)
n
=
x
2
f
n
(
x
2
,
y
2
)
+
y
2
f
n
(
y
2
,
x
2
)
{\displaystyle (x^{2}-y^{2})^{n}=x^{2}f_{n}(x^{2},y^{2})+y^{2}f_{n}(y^{2},x^{2})}
Which leads to the proposition by replacing
(
x
2
,
y
2
)
→
(
x
,
y
)
{\displaystyle (x^{2},y^{2})\rightarrow (x,y)}
Examples for (2):
(
x
−
y
)
3
=
x
(
x
+
3
y
)
2
−
y
(
y
+
3
x
)
2
(
x
+
y
)
3
=
x
(
x
−
3
y
)
2
+
y
(
y
−
3
x
)
2
{\displaystyle {\begin{array}{l}(x-y)^{3}=x(x+3y)^{2}-y(y+3x)^{2}\\(x+y)^{3}=x(x-3y)^{2}+y(y-3x)^{2}\end{array}}}
(
x
−
y
)
5
=
x
(
x
2
+
10
x
y
+
5
y
2
)
2
−
y
(
y
2
+
10
x
y
+
5
x
2
)
2
(
x
+
y
)
5
=
x
(
x
2
−
10
x
y
+
5
y
2
)
2
+
y
(
y
2
−
10
x
y
+
5
x
2
)
2
{\displaystyle {\begin{array}{l}(x-y)^{5}=x(x^{2}+10xy+5y^{2})^{2}-y(y^{2}+10xy+5x^{2})^{2}\\(x+y)^{5}=x(x^{2}-10xy+5y^{2})^{2}+y(y^{2}-10xy+5x^{2})^{2}\end{array}}}
Examples in
Z
{\displaystyle \mathbb {Z} }
:
17
3
=
(
5
+
12
)
3
=
5
×
31
2
+
12
×
3
2
17
5
=
(
5
+
12
)
5
=
5
×
145
2
+
12
×
331
2
17
7
=
(
5
+
12
)
7
=
5
×
6929
2
+
12
×
3767
2
17
9
=
(
5
+
12
)
9
=
5
×
138911
2
+
12
×
42921
2
{\displaystyle {\begin{array}{lll}17^{3}&=(5+12)^{3}=5\times 31^{2}&+12\times 3^{2}\\17^{5}&=(5+12)^{5}=5\times 145^{2}&+12\times 331^{2}\\17^{7}&=(5+12)^{7}=5\times 6929^{2}&+12\times 3767^{2}\\17^{9}&=(5+12)^{9}=5\times 138911^{2}&+12\times 42921^{2}\end{array}}}
Proposition
(
x
+
y
)
n
+
(
x
−
y
)
n
=
2
x
f
n
(
x
2
,
y
2
)
(
3
)
f
n
(
x
2
,
y
2
)
=
∑
k
=
0
n
−
1
(
x
+
y
)
n
−
1
−
k
(
y
−
x
)
k
(
4
)
{\displaystyle \quad {\begin{array}{lll}(x+y)^{n}+(x-y)^{n}&=2xf_{n}(x^{2},y^{2})&\quad (3)\\f_{n}(x^{2},y^{2})&=\sum _{k=0}^{n-1}(x+y)^{n-1-k}(y-x)^{k}&\quad (4)\end{array}}}
Proof
(1) implies (3)
(4):
(
x
+
y
)
n
−
(
x
−
y
)
n
=
2
y
f
n
(
y
2
,
x
2
)
u
n
−
v
n
=
(
u
−
v
)
(
∑
k
=
0
2
m
u
2
m
−
k
v
k
)
{\displaystyle {\begin{array}{clcl}(x+y)^{n}&-&(x-y)^{n}&=2yf_{n}(y^{2},x^{2})\\u^{n}&-&v^{n}&=(u-v)(\sum _{k=0}^{2m}u^{2m-k}v^{k})\end{array}}}
So
f
n
(
y
2
,
x
2
)
=
∑
k
=
0
2
m
(
x
+
y
)
2
m
−
k
(
x
−
y
)
k
{\displaystyle f_{n}(y^{2},x^{2})=\sum _{k=0}^{2m}(x+y)^{2m-k}(x-y)^{k}}
Let us consider a more detailed form of
f
n
(
x
,
y
)
{\displaystyle f_{n}(x,y)}
:
f
n
(
x
,
y
)
=
x
m
⏟
k=0
+
n
y
m
⏟
k=m
+
n
m
x
y
m
−
1
⏟
k=m-1
+
∑
k
=
1
m
−
2
(
n
2
k
)
x
m
−
k
y
k
{\displaystyle \quad f_{n}(x,y)=\underbrace {x^{m}} _{\text{k=0}}+\underbrace {ny^{m}} _{\text{k=m}}+\underbrace {nmxy^{m-1}} _{\text{k=m-1}}+\sum _{k=1}^{m-2}{n \choose 2k}x^{m-k}y^{k}}
Proposition
2
∣
x
y
,
x
∧
y
=
1
⇒
{
f
n
(
x
,
y
)
∧
f
n
(
y
,
x
)
=
1
(
5
)
n
∤
x
⇒
x
∧
f
n
(
x
,
y
)
=
1
(
6
)
n
∣
x
⇒
x
∧
f
n
(
x
,
y
)
=
n
(
7
)
{\displaystyle 2\mid xy,~x\wedge y=1\Rightarrow {\begin{cases}f_{n}(x,y)\wedge f_{n}(y,x)=1&(5)\\n\nmid x\Rightarrow x\wedge f_{n}(x,y)=1&(6)\\n\mid x\Rightarrow x\wedge f_{n}(x,y)=n&(7)\end{cases}}}
Proof
First,
f
n
(
x
,
y
)
=
x
m
+
n
y
m
+
x
y
P
(
x
,
y
)
{\displaystyle f_{n}(x,y)=x^{m}+ny^{m}+xyP(x,y)}
so
(
x
,
y
)
{\displaystyle (x,y)}
of opposite parity implies
f
n
(
x
,
y
)
{\displaystyle f_{n}(x,y)}
and
f
n
(
y
,
x
)
{\displaystyle f_{n}(y,x)}
odd.
The rule on gcd,
a
∧
b
=
a
∧
b
+
k
a
{\displaystyle a\wedge b=a\wedge b+ka}
, immediately implies (6) and (7).
Indeed,
f
n
(
x
,
y
)
=
x
m
+
y
P
(
x
,
y
)
{\displaystyle f_{n}(x,y)=x^{m}+yP(x,y)}
.
And for
n
∈
P
{\displaystyle n\in \mathbb {P} }
,
n
∣
(
n
2
k
)
{\displaystyle n\mid {n \choose 2k}}
so
f
n
(
n
x
,
y
)
n
=
y
m
+
x
P
(
x
,
y
)
{\displaystyle {\dfrac {f_{n}(nx,y)}{n}}=y^{m}+xP(x,y)}
Assertion (5) needs more attention.
Let us consider
p
{\displaystyle p}
a common odd prime divisor.
The second form gives us
(
x
−
y
)
n
≡
0
[
p
]
{\displaystyle (x-y)^{n}\equiv 0[p]}
, thus
x
≡
y
[
p
]
{\displaystyle x\equiv y[p]}
According to the definition of
f
n
{\displaystyle f_{n}}
f
n
(
x
,
y
)
≡
∑
k
=
0
m
(
n
2
k
)
x
m
−
k
y
k
≡
x
m
(
∑
k
=
0
m
(
n
2
k
)
)
≡
x
m
(
2
m
−
1
)
[
p
]
{\displaystyle f_{n}(x,y)\equiv \sum _{k=0}^{m}{n \choose 2k}x^{m-k}y^{k}\equiv x^{m}(\sum _{k=0}^{m}{n \choose 2k})\equiv x^{m}(2^{m-1})~[p]}
Thus
f
n
(
x
,
y
)
≡
0
[
p
]
⇒
x
≡
0
[
p
]
{\displaystyle f_{n}(x,y)\equiv 0[p]\Rightarrow x\equiv 0[p]}
, and the same
y
≡
0
[
p
]
{\displaystyle y\equiv 0[p]}
Every divisor of
f
n
(
x
,
y
)
{\displaystyle f_{n}(x,y)}
and
f
n
(
y
,
x
)
{\displaystyle f_{n}(y,x)}
divides
x
{\displaystyle x}
and
y
{\displaystyle y}
Propositions
n
∈
P
,
2
∣
x
y
,
x
∧
y
=
1
,
{
f
n
(
x
,
y
)
=
∏
p
i
v
i
⇒
p
i
≡
±
1
[
2
n
]
(
8
)
f
n
(
x
2
,
y
)
=
∏
p
i
v
i
⇒
p
i
≡
±
1
[
2
n
]
(
9
)
f
n
(
x
2
,
y
2
)
=
∏
p
i
v
i
⇒
p
i
≡
1
[
2
n
]
(
10
)
{\displaystyle n\in \mathbb {P} ,2\mid xy,~x\wedge y=1,\quad {\begin{cases}f_{n}(x,y)=\prod p_{i}^{v_{i}}&\Rightarrow p_{i}\equiv \pm 1[2n]&(8)\\f_{n}(x^{2},y)=\prod p_{i}^{v_{i}}&\Rightarrow p_{i}\equiv \pm 1[2n]&(9)\\f_{n}(x^{2},y^{2})=\prod p_{i}^{v_{i}}&\Rightarrow p_{i}\equiv \ 1[2n]&(10)\end{cases}}}
Proofs here for (10) p=±1[2n] and (8) p=1[2n] on math.stackexchange.com/
Note
Fermat theorem gives
f
n
(
x
,
y
)
≡
±
1
[
n
]
{\displaystyle f_{n}(x,y)\equiv \pm 1[n]}
and
f
n
(
x
2
,
y
)
≡
1
[
n
]
{\displaystyle f_{n}(x^{2},y)\equiv 1[n]}
. But what a surprise to discover that it also applies to all the prime factors! And much more specifically on the
f
n
(
x
2
,
y
2
)
{\displaystyle f_{n}(x^{2},y^{2})}
Let us remind the Fermat's theorem on sums of two squares:
x
2
+
y
2
=
∏
p
i
v
i
⇒
p
i
≡
1
[
4
]
{\displaystyle x^{2}+y^{2}=\prod p_{i}^{v_{i}}\Rightarrow p_{i}\equiv 1[4]}
And the Euler's theorem:
x
2
+
3
y
2
=
∏
p
i
v
i
⇒
p
i
≡
1
[
3
]
{\displaystyle x^{2}+3y^{2}=\prod p_{i}^{v_{i}}\Rightarrow p_{i}\equiv 1[3]}
, which is here
f
3
(
x
2
,
y
2
)
{\displaystyle f_{3}(x^{2},y^{2})}
Fermat had discovered that
2
n
−
1
{\displaystyle 2^{n}-1}
and
2
n
+
1
3
{\displaystyle {\dfrac {2^{n}+1}{3}}}
had
1
[
2
n
]
{\displaystyle 1[2n]}
prime factors (cf letters to Mersenne and Frenicle in 1640)
Let us note that these
f
n
(
x
2
,
y
2
)
{\displaystyle f_{n}(x^{2},y^{2})}
also appear in Fermat-Wiles theorem with (3)
Examples for
f
n
(
x
,
y
)
≡
±
1
[
2
n
]
{\displaystyle f_{n}(x,y)\equiv \pm 1[2n]}
f
5
(
6
,
11
)
=
1301
≡
1
[
10
]
f
7
(
6
,
11
)
=
181
×
239
≡
−
1
×
1
[
14
]
f
11
(
6
,
11
)
=
47820079
≡
1
[
22
]
f
13
(
6
,
11
)
=
8969
×
177269
≡
−
1
×
1
[
26
]
f
19
(
6
,
11
)
=
151
×
386989747169
≡
−
1
×
−
1
[
38
]
f
23
(
6
,
11
)
=
5278223
×
12238317893
≡
−
1
×
−
1
[
46
]
f
29
(
6
,
11
)
=
233
×
521
×
1309699
×
14932891583
≡
1
×
−
1
×
1
×
−
1
[
58
]
f
31
(
6
,
11
)
=
683
×
8803
×
13128774105430771
≡
1
×
−
1
×
1
[
62
]
.
.
.
{\displaystyle {\begin{array}{lll}f_{5}(6,11)&=1301&\equiv 1&[10]\\f_{7}(6,11)&=181\times 239&\equiv -1\times 1&[14]\\f_{11}(6,11)&=47820079&\equiv 1&[22]\\f_{13}(6,11)&=8969\times 177269&\equiv -1\times 1&[26]\\f_{19}(6,11)&=151\times 386989747169&\equiv -1\times -1&[38]\\f_{23}(6,11)&=5278223\times 12238317893&\equiv -1\times -1&[46]\\f_{29}(6,11)&=233\times 521\times 1309699\times 14932891583&\equiv 1\times -1\times 1\times -1&[58]\\f_{31}(6,11)&=683\times 8803\times 13128774105430771&\equiv 1\times -1\times 1&[62]\\...\end{array}}}
Examples for
f
n
(
x
2
,
y
)
≡
1
[
2
n
]
{\displaystyle f_{n}(x^{2},y)\equiv 1[2n]}
. The number of
−
1
[
n
]
{\displaystyle -1[n]}
factors is even
f
5
(
6
2
,
11
)
=
5861
≡
1
[
10
]
f
7
(
6
2
,
11
)
=
507809
≡
1
[
14
]
f
11
(
6
2
,
11
)
=
89
×
6029
×
7129
≡
1
×
1
×
1
[
22
]
f
13
(
6
2
,
11
)
=
883
×
2341
×
160627
≡
−
1
×
1
×
−
1
[
26
]
f
17
(
6
2
,
11
)
=
67
×
187067
×
199591969
≡
−
1
×
−
1
×
1
[
34
]
f
19
(
6
2
,
11
)
=
229
×
683
×
1901
×
2963
×
246469
≡
1
×
−
1
×
1
×
−
1
×
1
[
38
]
f
23
(
6
2
,
11
)
=
367
×
4457595074737380607
≡
−
1
×
−
1
[
46
]
f
29
(
6
2
,
11
)
=
1069838719517673460520580221
≡
1
[
58
]
f
31
(
6
2
,
11
)
=
92861463243343352659695472649
≡
1
[
62
]
.
.
.
{\displaystyle {\begin{array}{lll}f_{5}(6^{2},11)&=5861&\equiv 1&[10]\\f_{7}(6^{2},11)&=507809&\equiv 1&[14]\\f_{11}(6^{2},11)&=89\times 6029\times 7129&\equiv 1\times 1\times 1&[22]\\f_{13}(6^{2},11)&=883\times 2341\times 160627&\equiv -1\times 1\times -1&[26]\\f_{17}(6^{2},11)&=67\times 187067\times 199591969&\equiv -1\times -1\times 1&[34]\\f_{19}(6^{2},11)&=229\times 683\times 1901\times 2963\times 246469&\equiv 1\times -1\times 1\times -1\times 1&[38]\\f_{23}(6^{2},11)&=367\times 4457595074737380607&\equiv -1\times -1&[46]\\f_{29}(6^{2},11)&=1069838719517673460520580221&\equiv 1&[58]\\f_{31}(6^{2},11)&=92861463243343352659695472649&\equiv 1&[62]\\...\end{array}}}
Examples with both squared variables:
f
n
(
x
2
,
y
2
)
≡
1
[
2
n
]
{\displaystyle f_{n}(x^{2},y^{2})\equiv 1[2n]}
f
5
(
6
2
,
11
2
)
=
118061
≡
1
[
10
]
f
7
(
6
2
,
11
2
)
=
34188379
≡
1
[
14
]
f
11
(
6
2
,
11
2
)
=
23
×
89
×
2377
×
586961
≡
1
×
1
×
1
×
1
[
22
]
f
13
(
6
2
,
11
2
)
=
30187
×
27342279823
≡
1
×
1
[
26
]
f
19
(
6
2
,
11
2
)
=
2129
×
3079
×
3039225397425209
≡
1
×
1
×
1
[
38
]
f
23
(
6
2
,
11
2
)
=
1663964075070633572800332299
≡
1
[
46
]
f
29
(
6
2
,
11
2
)
=
59
×
11250493
×
60508154689065340703592763
≡
1
×
1
×
1
[
58
]
f
31
(
6
2
,
11
2
)
=
27466437659
×
422603394089373421296083801
≡
1
×
1
[
62
]
.
.
.
{\displaystyle {\begin{array}{lll}f_{5}(6^{2},11^{2})&=118061&\equiv 1&[10]\\f_{7}(6^{2},11^{2})&=34188379&\equiv 1&[14]\\f_{11}(6^{2},11^{2})&=23\times 89\times 2377\times 586961&\equiv 1\times 1\times 1\times 1&[22]\\f_{13}(6^{2},11^{2})&=30187\times 27342279823&\equiv 1\times 1&[26]\\f_{19}(6^{2},11^{2})&=2129\times 3079\times 3039225397425209&\equiv 1\times 1\times 1&[38]\\f_{23}(6^{2},11^{2})&=1663964075070633572800332299&\equiv 1&[46]\\f_{29}(6^{2},11^{2})&=59\times 11250493\times 60508154689065340703592763&\equiv 1\times 1\times 1&[58]\\f_{31}(6^{2},11^{2})&=27466437659\times 422603394089373421296083801&\equiv 1\times 1&[62]\\...\end{array}}}