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

 

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

TRIANGl SQR 1

 
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.

 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 

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About benvitalis

math grad - Interest: Number theory
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