设(n, m)=d, n=u*d, m=v*d, u, v是互素正整数
由题意t 被n/(m, n) 和 m/(m, n)整除,由于n/(m, n)和m/(m, n)互素,所以mn/(m, n)² | t
又由于 t=(m, t)*n/(m, n) 整除 mn/(m, n),所以设 t = mn/[d*(m, n)]
由mn/(m, n)² | t可得 d | (m, n)
代入 t/(m, t)= n/(m, n)可得 m/d = (m, t) = m/d * (d, n/(m, n)),所以d和n/(m, n)互素,同理d和m/(m, n)也互素
反过来,对所有满足d | (m, n) 并且d 与n/(m, n)和m/(m, n)互素的正整数d,t=mn/[d*(m, n)]都满足原方程组,所以是方程组的通解,d=1时t=[m, n]是其中一个特别的解
由题意t 被n/(m, n) 和 m/(m, n)整除,由于n/(m, n)和m/(m, n)互素,所以mn/(m, n)² | t
又由于 t=(m, t)*n/(m, n) 整除 mn/(m, n),所以设 t = mn/[d*(m, n)]
由mn/(m, n)² | t可得 d | (m, n)
代入 t/(m, t)= n/(m, n)可得 m/d = (m, t) = m/d * (d, n/(m, n)),所以d和n/(m, n)互素,同理d和m/(m, n)也互素
反过来,对所有满足d | (m, n) 并且d 与n/(m, n)和m/(m, n)互素的正整数d,t=mn/[d*(m, n)]都满足原方程组,所以是方程组的通解,d=1时t=[m, n]是其中一个特别的解