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

$1 \; + \; 2 \; + \; 3 \; + \; 4 \; + \; ... \; + \; x \; = \; y^2$

If   $(a, b)$   and   $(c, d)$   are two consecutive solutions of the Diophantine equation:
$1+2+3+4+...+x=y^2$    or    $x \, (x+1)/2 \; = \; y^2$
with   $a < c$
then    $c = 3 \, a + 4 \, b + 1$       $d = 2 \, a + 3 \, b + 1$

for example,   $(a, b) = (8, 6)$     [Line 2]
$c = (3)(8) + (4)(6) + 1 = 49$,
$d = (2)(8) + (3)(6) + 1 = 35$,
$(c, b) = (49, 35)$      [Line 3]

$T_{ \,3 \, a+4 \, b+1} \; = \; (2 \, a+3 \, b+1)^2$

Note that

$c \; - \; d \; = \; (3 \, a + 4 \, b + 1) \; - \; (2 \, a + 3 \, b + 1) \; = \; a \; + \; b$

$(a, b) = (8, 6)$      [Line 2]
$(c, d) = (49, 35)$      [Line 3]

$e = (d+b)^2 = (35+6)^2 = 41^2 = 1681$
the corresponding value,   $f = 1189 = 35^2 - 6^2 = d^2 - b^2$
$(e, f) = ( \,(d+b)^2, \; d^2-b^2 \,) = (1681, 1189)$      [Line 5]

Prove that if for any two consecutive solutions   $(a, b)$   and   $(c, d)$
$( \,(d+b)^2, \; d^2-b^2 \,)$   is a solution, too.

Also, prove that if   $(a,b)$,   $(c,d)$   and   $(e,f)$   are three consecutive solutions of the Diophantine equation, with   $a < c < e$
then   $(8 \, d^2, \; d \, (f-b))$   is a solution, too.