请教地里的前辈一道证明题

已知一个等式5n = 10x + 25y, n>=4, 并且x，y只能是非负整数，请用induction proof证明这个等式恒成立。我知道base case：n=4时， x=2，y=0，是成立的。Induction step：n=n+1时，5（n+1）=10x+25y+5，我不知道接下来该怎么证明。希望大家给点提示，谢谢。

我个人认为可以采用“跳跃归纳法”。

1. 设定两个base cases：n=4 和 n=5。

2. 假设 n=k 时等式成立。当 n=k+2 时，5(k+2)=10x+25y+10=10(x+1)+25y。

  5(n+1)
=10x+25y+5
=10(x-2)+25(y+1)     (a)
=10(x+3)+25(y-1)     (b)
x<2 and y<1 cannot both be true, therefore either (a) or (b) holds.

Not sure this is an induction proof:

1. You can easily find solution for n = 4, 5, 6 ... 13
2. For of n > 13, the result can be multiple of {n = 10} plus {n = 4, ... 13}

For example, n = 10,  solution is x = 0,  y = 2;    n = 6, solution is x = 3
Then for n = 1036, solution is y = 2 * 103, x = 3 .