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# 1.2 m, n  n3+ 1 mn − 1

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(1)

##    

 

  »º º

December 28, 2004

    

(2)

1     1

(3)

### 1

   

           

 d a, b  d | a, d | b d | am + bn,

 m, n  

 1.1  x99y  xy   x, y    x, y 



10x + 9 90 + y = x

y ⇒ 9xy − 90x + 9y = 0

⇒ xy − 10x + y = 0

⇒ (x + 1)(y − 10) = −10.



x + 1 2 5 10

y − 10 −5 −2 −1 ⇒ (x, y) = (1, 5), (4, 8),  (9, 9).



 1.2 m, n 

n3+ 1 mn − 1





 m, n 

 (mn − 1) | (n3+ 1), (mn − 1) | (mn − 1)

(mn − 1) | (n3+ 1)m − (mn − 1)n2 = n2+ m,

(mn − 1) | (n3+ 1)m3− (mn − 1)(m2n2+ mn + 1) = m3+ 1.



n3+ 1

mn − 1,m3+ 1

mn − 1,n2+ m mn − 1

  m, n   m ≥ n ≥ 1 m, n 

1 m ≤ 2(mn − 1) n2 ≤ mn ≤ 2(mn − 1) n2+ m

mn − 1 = 1, 2, 3,  4.

(1) 

n2+ m mn − 1 = 1,

 (n − 1)(m − n − 1) = 2 (m, n) = (5, 2), (5, 3) (2) 

n2+ m mn − 1 = 2,

 (2n − 1)(4m − 2n − 1) = 9 (m, n) = (2, 2), (3, 1)

(4)

(3) 

n2+ m mn − 1 = 3,

 (3n − 1)(9m − 3n − 1) = 28 (m, n) = (2, 1) (4) 

n2+ m mn − 1 = 4,

 (m, n) 

 

(m, n) = (2, 5), (3, 5), (1, 3), (1, 2).



(m, n) = (5, 2), (2, 5), (5, 3), (3, 5), (2, 2), (3, 1), (1, 3), (2, 1), (1, 2)

  

 1.1  ABC  15 P  BC    P A, P B, P C     P A 

 1.2  x66y  xy   x, y    x, y 

 1.3  x9999y  xy   x, y    x, y 

 1.4 

x2− y2 = 611

 x, y

 1.5 

(m2+ n)(m + n2) = (m + n)3

 m, n

 

 7    4   4  



p  

2p−1− 1 p

  

 



 p 

(5)

 - Ǳ

p  4  1  x, y 

x2− py2 = 4

   p   y   - Ǳ

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