## 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   $k$

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