The pentagonal numbers are the integers given byp5(n)=n(3n−1)/2 (n=0,1,2,…).Let (b,c,d) be one of the triples (1,1,2),(1,2,3),(1,2,6) and (2,3,4).We show that each n=0,1,2,… can be written as w+bx+cy+dz with w,x,y,z pentagonal numbers, which was first conjectured by Z.-W. Sun in 2016. In particular, any nonnegative integeris a sum of five pentagonal numbers two of which are equal; this refines a classical resultof Cauchy claimed by Fermat.
Citation: Dmitry Krachun, Zhi-Wei Sun. On sums of four pentagonal numbers with coefficients[J]. Electronic Research Archive, 2020, 28(1): 559-566. doi: 10.3934/era.2020029
[1] | Dmitry Krachun, Zhi-Wei Sun . On sums of four pentagonal numbers with coefficients. Electronic Research Archive, 2020, 28(1): 559-566. doi: 10.3934/era.2020029 |
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The pentagonal numbers are the integers given byp5(n)=n(3n−1)/2 (n=0,1,2,…).Let (b,c,d) be one of the triples (1,1,2),(1,2,3),(1,2,6) and (2,3,4).We show that each n=0,1,2,… can be written as w+bx+cy+dz with w,x,y,z pentagonal numbers, which was first conjectured by Z.-W. Sun in 2016. In particular, any nonnegative integeris a sum of five pentagonal numbers two of which are equal; this refines a classical resultof Cauchy claimed by Fermat.
For each
pm(n)=(m−2)(n2)+n (n∈N={0,1,2,…}). |
In particular, those
In 2016 Sun [9,Conjecture 5.2(ⅱ)] conjectured that each
p5(w)+bp5(x)+cp5(y)+dp5(z) with w,x,y,z∈N, |
provided that
(1,1,2),(1,2,2),(1,2,3),(1,2,4),(1,2,5),(1,2,6),(1,3,6),(2,2,4),(2,2,6),(2,3,4),(2,3,5),(2,3,7),(2,4,6),(2,4,7),(2,4,8). |
In 2017, Meng and Sun [5] confirmed this for
(b,c,d)=(1,1,2),(1,2,3),(1,2,6),(2,3,4). |
Theorem 1.1. Each
n=p5(x)+p5(y)+p5(z)+2p5(w). |
Remark 1.1. Clearly, Theorem 1.1 is stronger than the classical result that any nonnegative integer
is a sum of five pentagonal numbers. In Feb. 2019 the second author even conjectured that
any integer
Theorem 1.2. Any
Theorem 1.3. Let
We will prove Theorems 1.1-1.3 in Sections 2-4 respectively. Our proofs use some known results on ternary quadratic forms.
Those
{p5(x): x∈Z}= {n(3n−1)2: n∈N}∪{n(3n+1)2: n∈N}. |
Recently, Ju [3]
showed that for any positive integers
{a1p5(x1)+…+akpk(xk): x1,…,xk∈Z} |
contains all nonnegative integers whenever it contains the twelve numbers
1, 3, 8, 9, 11, 18, 19, 25, 27, 43, 98, 109. |
The generalized octagonal numbers are those
Lemma 2.1. Any positive even number
Proof. By Dickson [2,pp. 112-113],
N∖{x2+2y2+10z2: x,y,z∈Z}={8m+7: m∈N}∪{52k+1l: k,l∈N and l≡±1 (mod 5)}. |
Thus
0≡8n=s2+2(2w+z)2+10z2=(2r)2+12z2+8w(w+z) (mod 16), |
both
2n=r2+2(w+z2)2+10(z2)2=r2+(w+z)2+w2+2z2 =(x−y)2+(x+y)2+(x+y+z)2+2z2 =2x2+2y2+2z2+(x+y+z)2 |
and hence
Lemma 2.2. Let
23(n+B)+B−5B2=x2+y2+z2+(x+y+z)22≤B2. |
Then
Proof. Clearly,
x0=x+B, y0=y+B, z0=z+B, w0=w+B |
are nonnegative integers. Observe that
p5(x0)+p5(y0)+p5(z0)+2p5(w0)=3(x20+y20+z20+2w20)−(x0+y0+z0+2w0)2=3(5B2+x2+y2+z2+2w2)−5B2=2n+5B−5B2=n. |
This concludes the proof.
Proof of Theorem 1.1. We can easily verify the desired result for
√n3+16≤B≤√2n15+16, | (2.1) |
then
23(n+B)+B−5B2≥15(B−1/6)23+5B3−5B2=536>0 |
and
23(n+B)+B−5B2≤23(3(B−16))2+5B3−5B2 =B2−13(B−12)≤B2. |
Case 1.
As
n≥⌈32(√2/15−1/3)2⌉=8892, |
we have
√2n15+16−(√n3+16)=(√215−13)√n≥3 |
and hence there is an integer
0≤23(n+B)+B−5B2=2n+5B3−5B2≤B2. |
As the even number
23(n+B)+B−5B2=x2+y2+z2+(x+y+z)22. |
Now, by applying Lemma 2.2 we find that
Case 2.
In this case, we can easily verify the desired result when
n≥222289=⌈152(√2/15−1/3)2⌉. |
Choose
√2n15+16−(√n3+16)=(√215−13)√n≥15, |
there is an integer
23(n+B)+B−5B2=5(2q+B3−B2) |
and
2q+B3−B2≡−2q+B2−B2≡1−q−(B−1)2≡1−q−δ⧸≡0,±2 (mod 5). |
Thus, by applying Lemmas 2.1 and 2.2 we get that
In view of the above, we have completed the proof of Theorem 1.1.
Lemma 3.1. Let
6q=2x2+3y2+4z2+(2x+3y+4z)2. |
Proof. By K. Ono and K. Soundararajan [8], and Dickson [1], the Ramanujan form
x=a+b+2c, y=−b+2c, z=−3c, |
we have
2x2+3y2+4z2+(2x+3y+4z)2=6(a2+b2+10c2)=6q. |
This concludes the proof.
Lemma 3.2. Let
2n+10B3−10B2=2x2+3y2+4z2+(2x+3y+4z)2<(B+1)2. |
Then
Proof. Set
w0=w+B, x0=x+B, y0=y+B, z0=z+B |
are nonnegative integers. Observe that
p5(w0)+2p5(x0)+3p5(y0)+4p5(z0)=3(w20+2x20+3y20+4z20)−(w0+2x0+3y0+4z0)2=3(10B2+w2+2x2+3y2+4z2)−10B2=2n+10B−10B2=n. |
This ends the proof.
Proof of Theorem 1.2. We can verify the result for
n≥45325138=⌈(81−1/6+1/16)2(√1/15−√2/33)2⌉. |
Since
√n15+16−(√2n33+116)≥81, |
there is an integer
√2n33+116≤B≤√n15+16 |
such that
B≡−9n3+12n2−38n (mod 81) |
if
2n+10B3−10B2≥30(B−1/6)2+10B3−10B2=518>0 |
and
2n+10B3−10B2≤33(B−1/16)2+10B3−10B2=B2+4724B+11256<(B+1)2. |
Let
The proof of Theorem 1.2 is now complete.
Lemma 4.1. Let
q∈{7r: r∈Z and r≡1,2,4 (mod 7)}. |
Then there are
6q=x2+2y2+3z2+(x+2y+3z)2. |
Proof. Since
q∉{72k+1l: k,l∈N and l≡3,5,6 (mod 7)}, |
by [4,Theorem 2] we can write
x=6c, y=a−b−c, z=b−c, |
we have
x2+2y2+3z2+(x+2y+3z)2=6(a2+b2+7c2)=6q. |
This concludes the proof.
Lemma 4.2. Let
2n+(3δ+4)B3−(3δ+4)B2=x2+2y2+3δz2+(x+2y+3δz)2<(B+1)2. |
Then
Proof. Set
w0=w+B, x0=x+B, y0=y+B, z0=z+B |
are nonnegative integers. Observe that
p5(w0)+p5(x0)+2p5(y0)+3δp5(z0)=3(w20+x20+2y20+3δz20)−(w0+x0+2y0+3δz0)2=3((3δ+4)B2+w2+x2+2y2+3δz2)−(3δ+4)B2=2n+(3δ+4)B−(3δ+4)B2=n. |
This ends the proof.
Proof of Theorem 1.3 with
n≥808834881=⌈(7×81+1/48−1/6)2(√2/21−√1/12)2⌉. |
Since
√2n21+16−(√n12+148)≥7×81, |
there is an integer
√n12+148≤B≤√2n21+16 |
such that
0−12≡1−32≡6−02≡6 (mod 7),2−22≡5−02≡5 (mod 7), 3−02≡4−12≡3 (mod 7). |
Note that
2n+7B3−7B2≥21(B−1/6)2+7B3−7B2=736>0 |
and
2n+7B3−7B2≤24(B−1/48)2+7B3−7B2=B2+2B+1288<(B+1)2. |
It is easy to see that
q:=16(2n+7B3−7B2) |
is an integer divisible by
q7=16(2n0+B3−B2)≡−(9n0−6B3−B2)=(B+1)2−(3n0+1)≡1,2,4 (mod 7). |
By Lemma 4.1, we can write
Lemma 4.3. Let
q∉{52k+1l: k,l∈N and l≡±1 (mod 5)}. |
Then there are
6q=x2+2y2+6z2+(x+2y+6z)2. |
Proof. By Dickson [2,pp. 112-113], we can write
x=2a−b+3c, y=−a−b+3c, z=−2c, |
we have
x2+2y2+6z2+(x+2y+6z)2=6(a2+2b2+10c2)=6q. |
This ends the proof.
Proof of Theorem 1.3 with
n≥897099189=⌈(360+1/16−1/6)2(√1/15−√2/33)2⌉. |
Since
√n15+16−(√2n33+116)≥5×8×9, |
there is an integer
√2n33+116≤B≤√n15+16 |
such that
q=16(2n+10B3−10B2)=n+5B−15B29∈Z |
and
q5=n0+B−3B29≡3B2−B−n0≡B2−2B2−n0=(B−1)2−2n0−12⧸≡0,±1 (mod 5). |
As in the proof of Theorem 1.2, we also have
0<6q=2n+10B3−10B2<(B+1)2. |
Now applying Lemma 4.3 and Lemma 4.2 with
The proof of Theorem 1.3 with
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