A simple proof of Jacobi's four-square theorem

Jacobi's four-square theorem, which gives the number of representations of a positive integer as a sum of four squares, is shown to follow simply from the triple-product identity. 1. Recently [2], I showed how one can obtain Jacobi's two-square theorem from the triple-product identity. In this note I show that the triple-product identity also gives Jacobi's four-square theorem: Theorem 1. The number rA(n) of representations of the positive integer n as a sum of four squares is given by r4(«) = 8 £ d. <¡f|»,4+d 2. The triple product identity is CO n (1 + 1 -oo It follows easily (loc. cit.) that 00 (1) (a a-1) fi (1 a2x"){\ a" V')(l x") = £ (-i)'V +V"2 + ")/2. «Sl -oo Differentiate (1) with respect to a, put a — 1, divide by 2 and we obtain the identity (2) n(i-*")3=4i:(-i)"(2« + i)*<"2+">/2. »>1 Z -00 This is a celebrated identity of Jacobi [l, Theorem 357], and can be considered our starting point. Squaring (2) gives n0-*")6 = ! £ (-l)"' + "{2m + l)(2n + l)xim2 + "2 + m + n)/2. Received bv the editors April 15, 1986 and, in revised form, luly 15, 1986. 1980 Mathematics Subject Classification (1985 Revision). Primary 11E25, 11P05. ©1987 American Mathematical Society 0002-9939/87 $1.00 + $.25 per page 436 License or copyright restrictions may apply to redistribution; see https://www.ams.org/journal-terms-of-use PROOF OF JACOBI'S FOUR-SQUARE THEOREM 437 Now split the sum on the right into two, according as m + n is even or odd, to obtain n(l -*")6 = lf E (2m + \)(2n + l)x^2 + "2 + m+")/2 " » ! Ui s n (mod 2) £ (2w + 1)(2« +l)x""2 + "2 + m+")/2]. o»#n (mod 2) / In the first sum, set r = \(m + n), s = \{m n), and in the second set r = \{m — n — 1), s = t(w + h + 1), and, remarkably, the two sums coalesce to give IlO-*")6 = ̂ Ê ((2r + l)2-(2i)V+'2+'. Once again splitting the sum, we obtain i / 00 00 oc oc \ n(l-^")6 = 4 E ̂ E (2r+l)V2 + '-£ .v''2 +r E (2*)V) 2 w>l \ j——oo /-=-oo r~ — oo i""—oo = i I /xi + 44 E *'2+'E ^r2+rx4xf E *<2 dx dx v.v = -oc r — oo r = oc s = oo Making use of the triple-product identity, we obtain n(^-^")6 = y(n(i + ^2"-i)2(i-^2") n > 1 z I n > 1 1 +4x-^-)2n(l +x2")2(l x2") dx) bVi -2n(i + x2")2(i -x2«) n>l x4^^n (i + ^2"-i)2(i -x2")) dx „Vf yj Employing the product rule to evaluate the derivatives, we find nox"t = n (i + *2"-')2(i *2")(i + *2")2(i x2") n » 1 n» 1 L . ~ 2«jc2" . v 2«x2" — i i 2« " i _ 2n \ n » 1 x + ■* w » 1 1 ■* , n o + *2")2(i *a")o + *2"-i)2(i x2") n>l x|>£ ("-')"';-'-4£ w or, 1 _|_ 2n-l — 1 _ 2(( (1 > 1 1 + A » > 1 1 x no*")6 = n o + x2"_i)2(i + *2")2(i x2")2 í¡ > i « > i x(l-8El(2W"1)x2"" ^ 1 + x2""1 1 + x2" License or copyright restrictions may apply to redistribution; see https://www.ams.org/journal-terms-of-use 438 M. D. HIRSCHHORN Dividing both sides by n o + x")\i x"f = n o + *")2(i x2")2 n Ss 1 n > 1 we obtain = n a + x2""1)2^ + *2")2(i -x2")2, n»l I1"*")4-1 8£^(2w"1^2'"' 2nx2r „Vi U + x"} X\ \ 1 + *2"-1 1 + x2" Now, it is a simple consequence of the triple-product identity that n (££) -£(-i) v. so we have 00 / (I » 1 Putting -x for x, we obtain 1 + x2""1 1 + x2n £,-),.i+«s( 1 rcx" „ v4nx4" «>1 * x n>l *■ ~ x or. £*"* =i + 8£r nx i from which Theorem 1 follows directly. References 1. G. H. Hardy and E. M. Wright, An introduction to the theory of numbers, 4th ed., Clarendon Press, Oxford. 2. Michael D. Hirschhorn, A simple proof of Jacobi's two-square theorem. Amer. Math. Monthly 92 (1985), 579-580. School of Mathematics, University of New South Wales, P. O. Box 1, Kensington, New South Wales, Australia 2033 License or copyright restrictions may apply to redistribution; see https://www.ams.org/journal-terms-of-use