Square Numbers Satisfying $3x^2+x=4y^2+y$

[kaliprasad]

show that for x,y positive integers satisfying $3x^2+x= 4y^2+y$ each of x-y , 3x+3y+ 1 and 4x + 4y + 1 are squares. ( above equation has atleast one solution x= 30 and y = 26)

 

[Albert1]

show that for x,y positive integers satisfying $3x^2+x= 4y^2+y--(1)$
each of x-y =a^2 (2),
3x+3y+ 1=b^2 (3)and
4x + 4y + 1 (4)
are squares. ( above equation has atleast one solution x= 30 and y = 26)

Spoiler

from (2):$x-y=a^2>0---(*)$
from (3):$x+y=\dfrac{b^2-1}{3}=\dfrac{(b+1)(b-1)}{3}$
$\therefore (b+1)\,\, mod \,\ 3=0\,\, or \,\, (b-1)\,\, mod \,\, 3=0$
if $x+y \,\, even \,\, then \,\, b \,\, must \,\, be\,\ odd$
take $b=13$,we have $x+y=56$
if we let $x-y=a^2=4,\,\, and \,\ x+y=56$
the solution $x=30,y=26$ found
the solution of $a,b$ will meet the following equation:
$3a^2+6ab-b^2+1=0\,\, (a,b\in N)$
I wrote a computer program if b<1000000
the corresponding solution of :
[TABLE="width: 183"]
[TR]
[TD="width: 133, bgcolor: transparent"]x=$a^2+ab$

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[TD="width: 110, bgcolor: transparent"]y=$ab$

[/TD]
[/TR]
[TR]
[TD="bgcolor: transparent"]30
[/TD]
[TD="bgcolor: transparent"]26
[/TD]
[/TR]
[TR]
[TD="bgcolor: transparent"]5852
[/TD]
[TD="bgcolor: transparent"]5068
[/TD]
[/TR]
[TR]
[TD="bgcolor: transparent"]1135290
[/TD]
[TD="bgcolor: transparent"]983190
[/TD]
[/TR]
[TR]
[TD="bgcolor: transparent"]220240440
[/TD]
[TD="bgcolor: transparent"]190733816
[/TD]
[/TR]
[TR]
[TD="bgcolor: transparent"]42725510102
[/TD]
[TD="bgcolor: transparent"]37001377138
[/TD]
[/TR]
[/TABLE]

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[TD="width: 133, bgcolor: transparent, align: right"][/TD]
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[/TABLE]

 

[kaliprasad]

show that for x,y positive integers satisfying $3x^2+x= 4y^2+y$ each of x-y , 3x+3y+ 1 and 4x + 4y + 1 are squares. ( above equation has atleast one solution x= 30 and y = 26)

hint

Spoiler

$3x^2+x = 4y^2 + y$
or $3x^2- 3y^2 + x - y = y^2$
or $3(x+y)(x-y) + (x-y) = y^2$
or $(3x+3y+1)(x-y) = y^2$

 

[kaliprasad]

hint 2

Spoiler

$4x^2-4y^2 + x-y=x^2$
or $4(x-y)(x+y) + (x-y) = x^2$
or $(x-y)(4x+4y+1) = x^2$

 

[kaliprasad]

Spoiler

we have $3x^2+x = 4y^2 + y$
or $3x^2- 3y^2 + x - y = y^2$
or $3(x+y)(x-y) + (x-y) = y^2$
or $(3x+3y+1)(x-y) = y^2\cdots\, 1$also

$4x^2-4y^2 + x-y=x^2$
or $4(x-y)(x+y) + (x-y) = x^2$
or $(x-y)(4x+4y+1) = x^2\cdots\,2 $multiply (1) and (2) to get $(x-y)^2(3x+3y+1)(4x+4y+1)=x^2y^2$

or$(3x+3y+1)(4x+4y+1)=(\dfrac{xy}{x-y})^2$so $(3x+3y+1)(4x+4y+1)$ is a perfect squareas $4(3x+3y+1) - 3(4x+4y+1) = 1$ so $(3x+3y+1)$ and $(4x+4y+1)$ are coprime and hence perfect squares and the from (1) or (2) $(x-y)$ is a perfect square