system: [SYSTEM]

user: 4. 方程 φ 2 ( m ) = p 1 α 1 ⋅ p 2 α 2 ⋅ p 3 α 3 ⋅ ⋅ ⋅ p v α v 的解 本节将在前面两个定理的基础上，延申到更加一般的情况：即 φ 2 ( m ) = p 1 α 1 ⋅ p 2 α 2 ⋅ p 3 α 3 ⋅ ⋅ ⋅ p v α v (令 p 1 ≤ p 2 ≤ ⋅ ⋅ ⋅ ≤ p v 且均为质数)，由分析知，必有 p 1 = 2 。则 φ 2 ( m ) = 2 α 1 + 1 ⋅ p 2 α 2 ⋅ p 3 α 3 ⋅ ⋅ ⋅ p v α v ( α 1 ≥ − 1 )。由 m = q 1 β 1 ⋅ q 2 β 2 ⋅ ⋅ ⋅ q t β t ，其中 q 1 < q 2 < ⋯ < q t 且均为质数，则可得， q 1 β 1 − 1 ⋅ q 2 β 2 − 1 ⋅ ⋅ ⋅ q t β t − 1 ⋅ ( q 1 − 1 ) ⋅ ( q 2 − 1 ) ⋅ ⋅ ⋅ ( q t − 1 ) = 2 α 1 + 1 ⋅ p 2 α 2 ⋅ p 3 α 3 ⋅ ⋅ ⋅ p v α v (4.1) 其中必有 t ≤ v ，因为当 t > v 时，方程(4.1)无解。 4.1. 情况1 当 t = 1 时， q 1 β 1 − 1 ⋅ ( q 1 − 1 ) = 2 α 1 + 1 ⋅ p 2 α 2 ⋅ p 3 α 3 ⋅ ⋅ ⋅ p v α v (令 p ≤ q ) 情况1.1.1：当 β 1 = 1 时，由 q 1 − 1 = 2 α 1 + 1 ⋅ p 2 α 2 ⋅ p 3 α 3 ⋅ ⋅ ⋅ p v α v ，可得 m = q 1 ， q 1 = 2 s 1 1 ⋅ p 2 s 2 1 ⋅ p 3 s 3 1 ⋅ ⋅ ⋅ p v s v 1 + 1 (其中 s v i ≥ 0 )， i = [ 1 ， t ] ， α 1 = s 1 1 − 1 ， α j = s v i ， j = [ 2 ， t ] 。 情况1.1.2：当 β 1 ≥ 2 时，由 q 1 β 1 − 1 ⋅ ( q 1 − 1 ) = 2 α 1 + 1 ⋅ p 2 α 2 ⋅ p 3 α 3 ⋅ ⋅ ⋅ p v α v ，易知 q 1 = p u ， u ∈ [ 2 ， t ] ，可得 m = q 1 b 1 ， q 1 = 2 s 1 1 ⋅ p 2 s 2 1 ⋅ p 3 s 3 1 ⋅ ⋅ ⋅ p v s v 1 + 1 ， s u 1 (其中 s 1 1 ≥ − 1 ，且不包含 p u 这一素数)， α 1 = s 1 1 − 1 ， α j = s j 1 ， j = [ 2 ， t ] α u = s u 1 + b 1 − 1 ( s u 1 = 0 ) s u 1 之后也为0。 4.2. 情况2 当 t = 2 时， q 1 β 1 − 1 ⋅ ( q 1 − 1 ) ⋅ q 2 β 2 − 1 ⋅ ( q 1 − 1 ) = 2 α 1 + 1 ⋅ p 2 α 2 ⋅ p 3 α 3 ⋅ ⋅ ⋅ p v α v (令 p ≤ q ) 情况2.1.1：当 β 1 = 1 时，由 ( q 1 − 1 ) ⋅ ( q 2 − 1 ) = 2 α 1 + 1 ⋅ p 2 α 2 ⋅ p 3 α 3 ⋅ ⋅ ⋅ p v α v ，可得 m = q 1 q 2 ， q 1 = 2 s 1 1 ⋅ p 2 s 2 1 ⋅ p 3 s 3 1 ⋅ ⋅ ⋅ p v s v 1 + 1 ， q 2 = 2 s 1 2 ⋅ p 2 s 2 2 ⋅ p 3 s 3 2 ⋅ ⋅ ⋅ p v s v 2 + 1 ， α 1 = s 1 1 + s 1 2 − 1 ， α j = s 2 1 + s 2 2 ， j = [ 2 ， t ] 。 情况2.1.2：当 β 1 ≥ 2 ， β 2 = 1 时，有 q 1 β 1 − 1 ⋅ ( q 1 − 1 ) ⋅ ( q 2 − 1 ) = 2 α 1 + 1 ⋅ p 2 α 2 ⋅ p 3 α 3 ⋅ ⋅ ⋅ p v α v ，当 q 1 = p u 时， u ∈ [ 2 ， t ] ，可得 m = q 1 b 1 ⋅ q 2 ( b 1 ≥ 2 )， q 1 = 2 s 1 1 ⋅ p 2 s 2 1 ⋅ p 3 s 3 1 ⋅ ⋅ ⋅ p v s v 1 + 1 ， q 2 = 2 s 1 2 ⋅ p 2 s 2 2 ⋅ p 3 s 3 2 ⋅ ⋅ ⋅ p v s v 2 + 1 ， α 1 = s 1 1 + s 1 2 − 1 ， α j = s j 1 + s j 2 ，其中 j = [ 2 ， t ] ， α u = s u 1 + b 1 − 1 ( s u 1 = 0 ) s u 1 之后也为0。 情况2.1.3：当 β 1 ≥ 2 ， β 2 ≥ 2 时，由 q 1 β 1 − 1 ⋅ ( q 1 − 1 ) ⋅ q 2 β 2 − 1 ⋅ ( q 2 − 1 ) = 2 α 1 + 1 ⋅ p 2 α 2 ⋅ p 3 α 3 ⋅ ⋅ ⋅ p v α v ，易知 q 1 = p u 1 ， q 2 = p u 2 ，可得 m = q 1 b 1 ⋅ q 2 b 2 ， q 1 = 2 s 1 1 ⋅ p 2 s 2 1 ⋅ p 3 s 3 1 ⋅ ⋅ ⋅ p v s v 1 + 1 ， q 2 = 2 s 1 2 ⋅ p 2 s 2 2 ⋅ p 3 s 3 2 ⋅ ⋅ ⋅ p v s v 2 + 1 ， α 1 = s 1 1 + s 1 2 − 1 ， α j = s j 1 + s j 2 ， j = [ 2 ， t ] ， α u 1 = s u 1 + b 1 − 1 ( s u 1 = 0 ) s u 1 之后也为0， α u 2 = s u 2 + b 2 − 1 ( s u 1 = 0 ) s u 2 之后也为0。 4.3. 情况3 当 t ≥ 3 时， ∏ i = 1 t q i β i − 1 ⋅ ( q i − 1 ) = 2 α 1 + 1 ⋅ p 2 α 2 ⋅ p 3 α 3 ⋅ ⋅ ⋅ p v α v 。 情况3.1.1：当 β 1 = 1 时，由 ∏ i = 1 t ( q i − 1 ) = 2 α 1 + 1 ⋅ p 2 α 2 ⋅ p 3 α 3 ⋅ ⋅ ⋅ p v α v ，可得 m = q 1 q 2 ⋯ q t ， q i = 2 s i 1 ⋅ p 2 s i 1 ⋅ p 3 s i 1 ⋅ ⋅ ⋅ p v s i 1 + 1 ， α 1 = ∑ i = 1 t s i 1 − 1 ， α j = ∑ j = 1 v s j 1 ， j = [ 2 ， t ] 。 情况2.1.2：当 β 1 ≥ 2 ， β 2 = β 3 = ⋯ = β t = 1 时，有 q 1 β 1 − 1 ⋅ ∏ i = 1 t ( q 1 − 1 ) = 2 α 1 + 1 ⋅ p 2 α 2 ⋅ p 3 α 3 ⋅ ⋅ ⋅ p v α v ，当 q 1 = p u 时， u ∈ [ 2 ， t ] ，可得 m = q 1 b 1 ⋅ q 2 ⋅ ⋅ ⋅ q t ( b 1 ≥ 2 )， q i = 2 s i 1 ⋅ p 2 s i 1 ⋅ p 3 s i 1 ⋅ ⋅ ⋅ p v s i 1 + 1 ， α 1 = ∑ i = 1 t s i 1 − 1 ， α j = ∑ j = 1 v s j 1 ， j = [ 2 ， t ] ， α u = s u 1 + b 1 − 1 ( s u 1 = 0 )， s u 1 之后也为0。 情况2.1.3：当 β 1 ≥ 2 ， β 2 ≥ 2 ， β 3 = β 4 = ⋯ = β t = 1 时，由 q 1 β 1 − 1 ⋅ q 2 β 2 − 1 ⋅ ∏ i = 1 t ( q 1 − 1 ) = 2 α 1 + 1 ⋅ p 2 α 2 ⋅ p 3 α 3 ⋅ ⋅ ⋅ p v α v ，易知 q 1 = p u 1 ， q 2 = p u 2 ，可得 m = q 1 b 1 ⋅ q 2 b 2 ⋅ q 3 ⋅ ⋅ ⋅ q t ( b 1 ≥ 2 ， b 2 ≥ 2 )， q i = 2 s i 1 ⋅ p 2 s i 1 ⋅ p 3 s i 1 ⋅ ⋅ ⋅ p v s i 1 + 1 ， α 1 = ∑ i = 1 t s i 1 − 1 ， α j = s j 1 + s j 2 ， j = [ 2 ， t ] ， α u 1 = s u 1 + b 1 − 1 ( s u 1 = 0 ) s u 1 之后也为0。 α u 2 = s u 2 + b 2 − 1 ( s u 1 = 0 ) s u 2 之后也为0。 情况2.1.4：当 β 1 ≥ 2 ， β 2 ≥ 2 ， β 3 ≥ 2 ， β 4 = β 5 = ⋯ = β t = 1 时，由 q 1 β 1 − 1 ⋅ q 2 β 2 − 1 ⋅ q 3 β 3 − 1 ⋅ ∏ i = 1 t ( q 1 − 1 ) = 2 α 1 + 1 ⋅ p 2 α 2 ⋅ p 3 α 3 ⋅ ⋅ ⋅ p v α v ，易知 q 1 = p u 1 ， q 2 = p u 2 ， q 3 = p u 3 ，可得 m = q 1 b 1 ⋅ q 2 b 2 ⋅ q 3 b 3 ⋅ ⋅ ⋅ q t ( b 1 ≥ 2 ， b 2 ≥ 2 ， b 3 ≥ 2 )， q i = 2 s i 1 ⋅ p 2 s i 1 ⋅ p 3 s i 1 ⋅ ⋅ ⋅ p v s i 1 + 1 ， α 1 = ∑ i = 1 t s i 1 − 1 ， α j = s j 1 + s j 2 ， j = [ 2 ， t ] ， α u 1 = s u 1 + b 1 − 1 ( s u 1 = 0 )， s u 1 之后也为0。 α u 2 = s u 2 + b 2 − 1 ( s u 1 = 0 )， s u 2 之后也为0。 α u 3 = s u 3 + b 3 − 1 ( s u 1 = 0 ) s u 2 之后也为0。 由此可得，当 β i ≥ 2 时， m = q 1 b 1 ⋅ q 2 b 2 ⋅ q 3 b 3 ⋅ ⋅ ⋅ q t b t 。 将以上讨论概括为以下两个情况： 情况1： t = v ，当 q t = p v ( v = t + 1 )时， α 1 = β 1 − 2 ， q 1 = 2 ， α i = β i ( i = [ 2 ， v ] )，且 q i = 2 a i + 1 ( a i ≥ 0 )。 情况2： t > v ，必有 q t > p v ， q j = p j ， j = [ 1 ， u ] ，其中 u < v ，则 β u + 1 ， β u + 2 ， ⋯ ， β v = 1 ，且 q i = 2 s i 1 ⋅ p 2 s i 1 ⋅ p 3 s i 1 ⋅ ⋅ ⋅ p v s i 1 + 1 ， i = [ u + 1 ， v ] 。 于是定理3得证。