数论:求x3+x2+x=y2+y的正整数解.(数字为次数)

Question

数论:求x3+x2+x=y2+y的正整数解.(数字为次数)

Answer 1

x³=(y-x)(y+x+1),x,y为正整数
∴y>x,设y=x+a,a≥1.则x³=a(2x+a+1)
记d=(x,a)≥1,设a=md,x=nd,(m,n)=1
则n³d²=m(2nd+md+1),∵(d²,2nd+md+1)=1
∴d²|m,又(m,n)=1,∴m|d²,即m=d²
得n³=2nd+d³+1>d³,∴n>d,设n=d+b,b≥1
则b³+3b²d+3bd²=2d²+2bd+1
=> (3b-2)d²+(3b²-2b)d+b³-1=0
而由b≥1,得3b-2≥1,3b²-2b≥1,b³-1≥0
∴(3b-2)d²+(3b²-2b)d+b³-1≥d²+d>0,矛盾
所以原不定方程无正整数解.