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On sums of four pentagonal numbers with coefficients

  • The pentagonal numbers are the integers given byp5(n)=n(3n1)/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

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  • The pentagonal numbers are the integers given byp5(n)=n(3n1)/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 m=3,4,5,, the polygonal numbers of order m are given by

    pm(n)=(m2)(n2)+n  (nN={0,1,2,}).

    In particular, those p5(n) with nN are called pentagonal numbers. A famous claim of Fermat states that each nN can be written as a sum of m polygonal numbers of order m. This was proved by Lagrange for m=4 in 1770, by Gauss for m=3 in 1796, and by Cauchy for m5 in 1813. For Cauchy's polygonal number theorem, one may consult Nathanson [6] and [7,Chapter 1,pp. 3-34] for details. In 1830 Legendre refined Cauchy's polygonal number theorem by showing that for any m=5,6, every sufficiently large integer is a sum of five polygonal numbers of order m one of which is 0 or 1 (cf. [7,p. 33]).

    In 2016 Sun [9,Conjecture 5.2(ⅱ)] conjectured that each nN can be written as

    p5(w)+bp5(x)+cp5(y)+dp5(z)  with  w,x,y,zN,

    provided that (b,c,d) is among the following 15 triples:

    (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)=(1,2,2),(1,2,4). In this paper we prove the conjecture for

    (b,c,d)=(1,1,2),(1,2,3),(1,2,6),(2,3,4).

    Theorem 1.1. Each nN can be written as a sum of five pentagonal numbers two of which are equal, that is, there are x,y,z,wN such that

    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 n>33066 is a sum of three pentagonal numbers.

    Theorem 1.2. Any nN can be written as p5(w)+2p5(x)+3p5(y)+4p5(z) with w,x,y,zN.

    Theorem 1.3. Let δ{1,2}. Then any nN can be written as p5(w)+p5(x)+2p5(y)+3δp5(z) with w,x,y,zN.

    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(3x1)/2 with xZ are called generalized pentagonal numbers. Clearly,

    {p5(x): xZ}= {n(3n1)2: nN}{n(3n+1)2: nN}.

    Recently, Ju [3] showed that for any positive integers a1,,ak the set

    {a1p5(x1)++akpk(xk): x1,,xkZ}

    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 p8(x)=x(3x2) with xZ. In 2016, Sun [9] proved that any positive integer can be written as a sum of four generalized octagonal numbers one of which is odd. See also Sun [11] and [10] for representations of nonnegative integers in the form x(ax+b)/2+y(cy+d)/2+z(ez+f)/2 with x,y,z integers or nonnegative integers, where a,c,e are positive integers and b,d,f are integers with a+b,c+d,e+f all even.

    Lemma 2.1. Any positive even number n not in the set {52k+1m: k,mN and m±2 (mod 5)} can be written as x2+y2+z2+(x+y+z)2/2 with x,y,zZ.

    Proof. By Dickson [2,pp. 112-113],

        N{x2+2y2+10z2: x,y,zZ}={8m+7: mN}{52k+1l: k,lN and l±1 (mod 5)}.

    Thus 8n=s2+2t2+10z2 for some s,t,zZ. Clearly, 2s and tz (mod 2). Without loss of generality, we may assume that tz (mod 4) if 2z. (If tz (mod 4) with z odd, then tz (mod 4).) Write s=2r and t=2w+z with r,wZ. Then 2w if 2z. Since

    08n=s2+2(2w+z)2+10z2=(2r)2+12z2+8w(w+z) (mod 16),

    both rz and w(w+z) are even. If 2z then 2w. Recall that 2w if 2z. So wzr (mod 2). Now, both x=(r+w)/2 and y=(wr)/2 are integers. Observe that

    2n=r2+2(w+z2)2+10(z2)2=r2+(w+z)2+w2+2z2      =(xy)2+(x+y)2+(x+y+z)2+2z2    =2x2+2y2+2z2+(x+y+z)2

    and hence n=x2+y2+z2+(x+y+z)2/2. This ends the proof.

    Lemma 2.2. Let nN. Suppose that there are BN and x,y,zZ such that 3n+B and

    23(n+B)+B5B2=x2+y2+z2+(x+y+z)22B2.

    Then n=p5(x0)+p5(y0)+p5(z0)+2p5(w0) for some x0,y0,z0,w0N.

    Proof. Clearly, w=(x+y+z)/2Z. As |x|,|y|,|z|,|w|B, all the numbers

    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+5B5B2=n.

    This concludes the proof.

    Proof of Theorem 1.1. We can easily verify the desired result for n=0,,8891. Below we assume that n8892. If

    n3+16B2n15+16, (2.1)

    then

    23(n+B)+B5B215(B1/6)23+5B35B2=536>0

    and

    23(n+B)+B5B223(3(B16))2+5B35B2                            =B213(B12)B2.

    Case 1. 5n.

    As

    n32(2/151/3)2=8892,

    we have

    2n15+16(n3+16)=(21513)n3

    and hence there is an integer B satisfying (2.1) with Bn (mod 3). By the above,

    023(n+B)+B5B2=2n+5B35B2B2.

    As the even number 23(n+B)+B5B2 is not divisible by 5, in light of Lemma 2.1 there are x,y,zZ such that

    23(n+B)+B5B2=x2+y2+z2+(x+y+z)22.

    Now, by applying Lemma 2.2 we find that n=p5(x0)+p5(y0)+p5(z0)+2p5(w0) for some x0,y,z0,w0N.

    Case 2. n=5q for some qN.

    In this case, we can easily verify the desired result when 8892n222288. Below we assume that

    n222289=152(2/151/3)2.

    Choose δ{0,±1} such that 1qδ0,±2 (mod 5). As

    2n15+16(n3+16)=(21513)n15,

    there is an integer B satisfying (2.1) such that Bn (mod 3) and (B1)2δ (mod 5). Note that

    23(n+B)+B5B2=5(2q+B3B2)

    and

    2q+B3B22q+B2B21q(B1)21qδ0,±2 (mod 5).

    Thus, by applying Lemmas 2.1 and 2.2 we get that n=p5(x0)+p5(y0)+p5(z0)+2p5(w0) for some x0,y,z0,w0N.

    In view of the above, we have completed the proof of Theorem 1.1.

    Lemma 3.1. Let qN with q odd and not squarefree, or 2q and q{4k(16l+6): k,lN}. Then there are x,y,zZ such that

    6q=2x2+3y2+4z2+(2x+3y+4z)2.

    Proof. By K. Ono and K. Soundararajan [8], and Dickson [1], the Ramanujan form x2+y2+10z2 represents q. Write q=a2+b2+10c2 with a,b,cZ. Then, for

    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 nN. Suppose that there are BN and x,y,zZ such that

    2n+10B310B2=2x2+3y2+4z2+(2x+3y+4z)2<(B+1)2.

    Then n=p5(w0)+2p5(x0)+3p5(y0)+4p5(z0) for some w0,x0,y0,z0N.

    Proof. Set w=(2x+3y+4z). As |w|,|x|,|y|,|z|B, all the numbers

    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+10B10B2=n.

    This ends the proof.

    Proof of Theorem 1.2. We can verify the result for n=0,,45325137 directly via a computer. Below we assume that

    n45325138=(811/6+1/16)2(1/152/33)2.

    Since

    n15+16(2n33+116)81,

    there is an integer B with

    2n33+116Bn15+16

    such that

    B9n3+12n238n (mod 81)

    if n is odd, and B3n1 (mod 8) and B3n22n (mod 9) if n is even. Note that

    2n+10B310B230(B1/6)2+10B310B2=518>0

    and

    2n+10B310B233(B1/16)2+10B310B2=B2+4724B+11256<(B+1)2.

    Let q=(n+5B15B2)/9. When n is odd, we can easily see that q is an odd integer divisible by 9. When n is even, q is an even integer with q4 (mod 8), and hence q4k(16l+6) for any k,lN. By Lemma 3.1, we can write 6q=(2n+10B)/310B2 as 2x2+3y2+4z2+(2x+3y+4z)2 with x,y,zZ. Applying Lemma 3.2, we see that n=p5(w0)+2p5(x0)+3p5(y0)+4p5(z0) for some w0,x0,y0,z0N.

    The proof of Theorem 1.2 is now complete.

    Lemma 4.1. Let qN be a multiple of 9 with 7q or

    q{7r: rZ and r1,2,4 (mod 7)}.

    Then there are x,y,zZ such that

    6q=x2+2y2+3z2+(x+2y+3z)2.

    Proof. Since 9q and

    q{72k+1l: k,lN and l3,5,6 (mod 7)},

    by [4,Theorem 2] we can write q as a2+b2+7c2 with a,b,cZ. For

    x=6c, y=abc, z=bc,

    we have

    x2+2y2+3z2+(x+2y+3z)2=6(a2+b2+7c2)=6q.

    This concludes the proof.

    Lemma 4.2. Let nN and δ{1,2}. Suppose that there are BN and x,y,zZ such that

    2n+(3δ+4)B3(3δ+4)B2=x2+2y2+3δz2+(x+2y+3δz)2<(B+1)2.

    Then n=p5(w0)+p5(x0)+2p5(y0)+3δp5(z0) for some w0,x0,y0,z0N.

    Proof. Set w=(x+2y+3δz). As |w|,|x|,|y|,|z|B, all the numbers

    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 δ=1. We can verify the desired result for n=0,1,,808834880 directly via a computer. Below we assume that

    n808834881=(7×81+1/481/6)2(2/211/12)2.

    Since

    2n21+16(n12+148)7×81,

    there is an integer B with

    n12+148B2n21+16

    such that B18n3+3n235n (mod 81), and 3n/7+1(B+1)23,5,6 (mod 7) if 7n. Such an integer B exists in view of the Chinese Remainder Theorem and the simple observations

    0121326026 (mod 7),2225025 (mod 7), 3024123 (mod 7).

    Note that

    2n+7B37B221(B1/6)2+7B37B2=736>0

    and

    2n+7B37B224(B1/48)2+7B37B2=B2+2B+1288<(B+1)2.

    It is easy to see that

    q:=16(2n+7B37B2)

    is an integer divisible by 9. If n=7n0 for some n0N, then

    q7=16(2n0+B3B2)(9n06B3B2)=(B+1)2(3n0+1)1,2,4 (mod 7).

    By Lemma 4.1, we can write 6q=(2n+7B)/37B2 as x2+2y2+3z2+(x+2y+3z)2 with x,y,zZ. Applying Lemma 4.2 with δ=1, we see that n=p5(w0)+p5(x0)+2p5(y0)+3p5(z0) for some w0,x0,y0,z0N. This completes the proof.

    Lemma 4.3. Let qN with q7 (mod 8) or

    q{52k+1l: k,lN and l±1 (mod 5)}.

    Then there are x,y,zZ such that

    6q=x2+2y2+6z2+(x+2y+6z)2.

    Proof. By Dickson [2,pp. 112-113], we can write q as a2+2b2+10c2 with a,b,cZ. For

    x=2ab+3c, y=ab+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 δ=2. We can verify the desired result for n=0,1,,897099188 directly via a computer. Below we assume that

    n897099189=(360+1/161/6)2(1/152/33)2.

    Since

    n15+16(2n33+116)5×8×9,

    there is an integer B with

    2n33+116Bn15+16

    such that B3n22n (mod 9) and Bn2n1 (mod 8), and (B1)22n0±1,2n02 (mod 5) if n=5n0 with n0N. Then

    q=16(2n+10B310B2)=n+5B15B29Z

    and q7 (mod 8). If n=5n0 for some n0N, then

    q5=n0+B3B293B2Bn0B22B2n0=(B1)22n0120,±1 (mod 5).

    As in the proof of Theorem 1.2, we also have

    0<6q=2n+10B310B2<(B+1)2.

    Now applying Lemma 4.3 and Lemma 4.2 with δ=2, we obtain that n=p5(w0)+p5(x0)+2p5(y0)+6p5(z0) for some w0,x0,y0,z0N.

    The proof of Theorem 1.3 with δ=2 is now complete.



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