(2) Assuming that n=k(k≥8), there exists x, y∈N, so that k=3x+5y.

(1) If y=0, k≥8 and x≥3 show that k+1= 3x+1= 3 (x-3)+10 = 3 (x-3)+5× 2 holds.

② If y≠0, then k+1= 3x+5y+1= 3x+5 (y-1)+6 = 3 (x+2)+5 (y-1) holds;

That is to say, when n=k+ 1, the proposition also holds.

From (1)(2), the proposition holds.