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

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The equation $3x^2 + x = 4y^2 + y$ has positive integer solutions, with one known solution being x = 30 and y = 26. For these values, it is demonstrated that the expressions x - y, 3x + 3y + 1, and 4x + 4y + 1 are all perfect squares. The discussion emphasizes the need to prove that these expressions yield square numbers for any positive integer solutions to the equation. The mathematical relationships and properties of squares are central to the analysis. This exploration contributes to a deeper understanding of the equation's solutions and their implications.
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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)
 
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kaliprasad said:
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)
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=$a^2+ab$

[/TD]
[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]
[TABLE="width: 195"]
[TR]
[TD="width: 127, bgcolor: transparent, align: right"][/TD]
[TD="width: 133, bgcolor: transparent, align: right"][/TD]
[/TR]
[TR]
[TD="bgcolor: transparent, align: right"][/TD]
[TD="bgcolor: transparent, align: right"][/TD]
[/TR]
[TR]
[TD="bgcolor: transparent, align: right"][/TD]
[TD="bgcolor: transparent, align: right"][/TD]
[/TR]
[TR]
[TD="bgcolor: transparent, align: right"][/TD]
[TD="bgcolor: transparent, align: right"][/TD]
[/TR]
[TR]
[TD="bgcolor: transparent, align: right"][/TD]
[TD="bgcolor: transparent, align: right"][/TD]
[/TR]
[/TABLE]
 
Last edited:
kaliprasad said:
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

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

$4x^2-4y^2 + x-y=x^2$
or $4(x-y)(x+y) + (x-y) = x^2$
or $(x-y)(4x+4y+1) = x^2$
 
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
 
Here is a little puzzle from the book 100 Geometric Games by Pierre Berloquin. The side of a small square is one meter long and the side of a larger square one and a half meters long. One vertex of the large square is at the center of the small square. The side of the large square cuts two sides of the small square into one- third parts and two-thirds parts. What is the area where the squares overlap?

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