%0 Journal Article %T 不定方程<i>x</i>(<i>x</i>+1)(<i>x</i>+2)(<i>x</i>+<i>3</i>)=33<i>y</i>(<i>y</i>+1)(<i>y</i>+2)(<i>y</i>+3)的整数解的研究<br>A Research of the Integer Solution of the Diophantine Equation <i>x</i>(<i>x</i>+1)(<i>x</i>+2)(<i>x</i>+<i>3</i>)=33<i>y</i>(<i>y</i>+1)(<i>y</i>+2)(<i>y</i>+3) %A 陈琼 %J 西南大学学报(自然科学版) %D 2018 %R 10.13718/j.cnki.xdzk.2018.04.006 %X $主要运用Pell方程、递推序列、同余式及(非)平方剩余等一些初等的证明方法,对不定方程 $ x(x + 1)(x + 2)(x + 3) = 33y(y + 1)(y + 2)(y + 3) $ 的解进行了研究.证明了该不定方程仅有1组正整数解(<i>x</i>,<i>y</i>)=(9,3).同时给出了不定方程(<i>x</i><sup>2</sup>+3<i>x</i>+1)<sup>2</sup>-33<i>y</i><sup>2</sup>=-32的全部整数解.$<br>$In this paper, with such elementary methods as Pell equation, recurrence sequence, congruent form and quadratic (non-)residue, the author studies the diophantine equation $ x(x + 1)(x + 2)(x + 3) = 33y(y + 1)(y + 2)(y + 3) $ and shows that its only solution in positive integers is (<i>x</i>, <i>y</i>))=(9, 3). She also obtains all the integer solutions of the diophantine equation (<i>x</i><sup>2</sup>+3<i>x</i>+1)<sup>2</sup>-33<i>y</i><sup>2</sup>=-32. %K 不定方程 %K 整数解 %K 递归数列 %K 平方剩余< %K br> %K diophantine equation %K integer solution %K recurrence sequence %K quadratic remainder %U http://xbgjxt.swu.edu.cn/jsuns/html/jsuns/2018/4/201804006.htm