Diophantine equation: 1 + 2 + 3 + … + x = y^2 — Part 2

 
 
Diophantine equation:   1 + 2 + 3 + … + x = y^2   — Part 1

 

1 \; + \; 2 \; + \; 3 \; + \; ... \; + \; x \; = \; y^2
 
The first few solutions:

  (x_1, \; y_1) \; = \; (1, \; 1)
  (x_2, \; y_2) \; = \; (8, \; 6)
  (x_3, \; y_3) \; = \; (49, \; 35)
  (x_4, \; y_4) \; = \; (288, \; 204)
  (x_5, \; y_5) \; = \; (1681, \; 1189)
  (x_6, \; y_6) \; = \; (9800, \; 6930)
  (x_7, \; y_7) \; = \; (57121, \; 40391)
  (x_8, \; y_8) \; = \; (332928, \; 235416)
  (x_9, \; y_9) \; = \; (1940449, \; 1372105)
(x_{10}, \; y_{10}) \; = \; (11309768, \; 7997214)
(x_{11}, \; y_{11}) \; = \; (65918161, \; 46611179)
(x_{12}, \; y_{12}) \; = \; (384199200, \; 271669860)
(x_{13}, \; y_{13}) \; = \; (2239277041, \; 1583407981)
(x_{14}, \; y_{14}) \; = \; (13051463048, \; 9228778026)
(x_{15}, \; y_{15}) \; = \; (76069501249, \; 53789260175)
(x_{16}, \; y_{16}) \; = \; (443365544448, \; 313506783024)
(x_{17}, \; y_{17}) \; = \; (2584123765441, \; 1827251437969)
(x_{18}, \; y_{18}) \; = \; (15061377048200, \; 10650001844790)
(x_{19}, \; y_{19}) \; = \; (87784138523761, \; 62072759630771)
(x_{20}, \; y_{20}) \; = \; (511643454094368, \; 361786555939836)
(x_{21}, \; y_{21}) \; = \; (2982076586042449, \; 2108646576008245)
(x_{22}, \; y_{22}) \; = \; (17380816062160328, \; 12290092900109634)
(x_{23}, \; y_{23}) \; = \; (101302819786919521, \; 71631910824649559)

 
(y_1) \,(y_3) + 1 \; = \; (1)(35) + 1 \; = \; 36 \; = \; 6^2 \; = \; ( \,y_2 \,)^2
(y_2) \,(y_4) + 1 \; = \; (6)(204) + 1 \; = 1225 \; = \; 35^2 \; = \; ( \,y_3 \,)^2
(y_3) \,(y_5) + 1 \; = \; (35)(1189) + 1 \; = \; 41616 \; = \; 204^2 \; = \; ( \,y_4 \,)^2
 

Show that for any   k

( \,y_{k-1} \,) \,( \,y_{k+1} \,) \; + \; 1 \; = \; ( \,y_{k} \,)^2
 
 

 
 
 
 
 

 
 
 
 
 
 
 
 
 
 
 
 
 
 
 

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About benvitalis

math grad - Interest: Number theory
This entry was posted in Number Puzzles and tagged . Bookmark the permalink.

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