$B9)A!Bg$NDMK\$G$9(B.

In article <c4e88bc5-0de6-4b8d-ab61-e2a9aa6623d8@39g2000yqa.googlegroups.com>
KyokoYoshida <kyokoyoshida123@gmail.com> writes:
> $B2<5-$N$h$&$K(B
> http://beauty.geocities.jp/yuka26076/study/Number_Theory/prop151_955_p137.JPG
> $B6qBNE*$K2r$r9=@.$;$:$K2r$NB8:_$NJ]>Z$r>ZL@$7$F$_$?$N$G$9$,(B
> $B$3$l$O4V0c$$$G$7$g$&$+(B?

$B$=$l$b(B k $B$K$D$$$F$N(B induction $B$G(B g_k(x), h_k(x) $B$r(B
$B!V9=@.!W$7$F$k$o$1$G(B, $BF1$8$3$H$G$9(B.
 g_k(x) = x - b_k $B$N(B b_k $B$OA4$/F1$8$G$9$M(B.

> Now we shall show that x_0+5x_1+5^2x_2+5^3x_3+ \cdots +5^{k-1}x_{k-1} is a
> solution of [0].

Now we shall show that there is a solution b_k of [0] of the form
 b_k = x_0 + 5 x_1 + 5^2 x_3 + \cdots + 5^{k-1} x_{k-1},
where each x_k (0 \leq k \leq k-1) is one of the integers { 0, 1, 2, 3, 4 }.

> Letting b_k$B":(BS_k$B!D!Z(B1$B![(B, we can write b_k$B":(B(x_0+5x_1+5^2x_2+5^3x_3+$B!D(B+5^{k-1}
> x_{k-1})mod5^k$B">(BS_k$B!D!Z(B2$B![(B
> (where x_k$Bch$(D??(B1,2,3,4})
> $B$H$7$F?J$a$F$$$1$P$$$$$N$G$9$M!#(B

$B>e$r=q$1$P2<$OMW$i$J$$$G$7$g$&(B.

> 2 x_k $B"a(B 4 $B"a(B -1(mod 5)$B$5$(J,$+$l$P(B(2 x_k)^{-1} -1(mod 5)$B$r;H$o$:$K(B
> http://beauty.geocities.jp/yuka26076/study/Number_Theory/example5_39_vol11.JPG
> $B$H$G$-$^$9$M!#(B

$B$=$&(B, $B:G=i$N(B <100927172650.M0206377@ras2.kit.ac.jp> $B$G$b(B
 <101012174820.M0104817@ras2.kit.ac.jp> $B$G$b(B
 x_k $B$O$=$&







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