Show that x3+x+a2=y2 has at least one pair of positive integer solution (x,y) for each positive integer a.
Problem
Source: North Korea Team Selection Test 2013 #6
Tags: induction, number theory proposed, number theory
17.05.2014 17:56
Since x(x2+1)=y2−a2=(y+a)(y−a), it is sufficient to show the existence of positive numbers b,c,d,e, which satisfy y+a=bc,y−a=de,x=bd,x2+1=ce,bc−de=2a,ce−(bd)2=1. We propose the following lemma: The sequence un is defined as follows: u∈N,u0=0,u1=1,un+2=uun+1+un where n=0,1,2,3.... Then unun+2−u2n+1=(−1)n+1,n≥0. The proof of lemma: We prove by the induction on n. For n=0, it holds true. Assume it holds true for n−1(≥0). Then unun+2−u2n+1=un(uun+1+un)−u2n+1=−un+1(un+1−uun)+u2n=−(un+1un−1−u2n)=(−1)n+1. Now, let c=u3=u2+1,e=u5=u4+3u2+1,bd=u4=u(u2+2), then from the lemma we have ce−(bd)2=1. Let d=2a,u=4a2. Then b=2a(u2+2) and bc−de=2a(u2+2)(u2+1)−2a(u4+3u2+1)=2a. As a result, x=bd=4a2(16a4+2)y=2a(16a4+2)(16a4+1)−a is one pair of positive integer solution of the equation.
01.07.2021 06:49
Fun note: let P=(0,a) be the obvious integral point on the elliptic curve y2=x3+x+a2,then [3]P=(8a2(8a4+1),4a(8a4+1)(16a4+1)−a) is the integral point found above ([3]P=P⊕P⊕P, where ⊕ is the addition on elliptic curves). This is also true for the elliptic curve y2=x3+2x+a2. In particular, let P=(0,a) be the obvious integral point, then [3]P=(4a2(a4+1),a(8a8+12a4+3)) is another integral point on the curve.