已知正整数x和y满足2x²-1＝y∧15 证明,若x＞1,则x可被5整除

应该是2005年俄罗斯数学奥林匹克10年级第7题.
设t = y^5,将方程写为2x² = 1+t³ = (1+t)(1-t+t²).
由1-t+t² = (1+t)(t-2)+3,可知(1+t,1-t+t²) = 1或3.
若(1+t,1-t+t²) = 1,由(1+t)(1-t+t²) = 2x²且1-t+t²为奇数可知1-t+t²是完全平方数.
但由x > 1,有t > 1,(t-1)² 因此(1+t,1-t+t²) = 3,3 | 1+t = 1+y^5,可知y ≡ -1 (mod 3).
设s = y³,将方程写为2x² = 1+s^5 = (1+s)(1-s+s²-s³+s^4).
由1-s+s²-s³+s^4 = (1+s)(s³-2s²+3s-4)+5,可知(1+s,1-s+s²-s³+s^4) = 1或5.
若(1+s,1-s+s²-s³+s^4) = 1,由(1+s)(1-s+s²-s³+s^4) = 2x²且1-s+s²-s³+s^4为奇数,
可知1-s+s²-s³+s^4是完全平方数.
然而,由y ≡ -1 (mod 3),s = y³ ≡ -1 (mod 3),代入得1-s+s²-s³+s^4 ≡ 5 ≡ -1 (mod 3),
不可能为完全平方数,矛盾.
因此(1+s,1-s+s²-s³+s^4) = 5,有5 | s+1 | 2x²,故5 | x,所证结论成立.