试求出所有整数n,使得代数式2n2+n-29的值是某两个连续自然数的平方和

Question

试求出所有整数n,使得代数式2n2+n-29的值是
某两个连续自然数的平方和

Answer 1

相当于求2n²+n-29 = m²+(m+1)² = 2m²+2m+1的整数解.
也即2n²+n = 2m²+2m+30.
乘以8并配方为(4n+1)²-1 = (4m+2)²+236,即(4n+1)²-(4m+2)² = 237.
于是(4n-4m-1)(4n+4m+3) = 237,4n+4m+3与4n-4m-1均为237的约数.
237 = 3·79,约数有±1,±3,±79,±237.
又4n+4m+3 > 4n-4m-1 (m为自然数),且二者都除以4余3.
只需考虑4n+4m+3 = -1,79,相应4n-4m-1 = -237,3.
分别解得n = -30,m = 29与n = 10,m = 9.
可验证n = -30时2n²+n-29 = 1741 = 29²+30².
n = 10时2n²+n-29 = 181 = 9²+10².
故n的所有可能取值为-30与10.