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

In summary, square numbers are numbers that can be expressed as the product of two equal integers. To determine if a number is a square number, we can check if its square root is a whole number. The equation $3x^2+x=4y^2+y$ can be rewritten in terms of square numbers as $(3x+1)^2=(2y+1)^2$, making it easier to solve for all possible solutions. This equation is significant as it is an example of a Diophantine equation and has a unique property that simplifies the solving process.
  • #1
kaliprasad
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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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  • #2
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 :
x=$a^2+ab$

y=$ab$

30
26
5852
5068
1135290
983190
220240440
190733816
42725510102
37001377138
 
Last edited:
  • #3
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 $(3x+3y+1)(x-y) = y^2$
 
  • #4
hint 2

$4x^2-4y^2 + x-y=x^2$
or $4(x-y)(x+y) + (x-y) = x^2$
or $(x-y)(4x+4y+1) = x^2$
 
  • #5
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
 

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

1. What are square numbers?

Square numbers are numbers that can be expressed as the product of two equal integers. For example, 4 is a square number because it can be expressed as 2x2.

2. How do you determine if a number is a square number?

A number is a square number if its square root is a whole number. In other words, if the square root of a number can be expressed as an integer, then that number is a square number.

3. How can you rewrite the equation $3x^2+x=4y^2+y$ in terms of square numbers?

We can rewrite the equation as $3x^2+3x+1=4y^2+4y+1$, which is equivalent to $(3x+1)^2=(2y+1)^2$. This means that the left side is a square number and the right side is a square number, so solving this equation will give us all the possible solutions.

4. How can you solve the equation $3x^2+x=4y^2+y$?

We can use the above rewrite to solve the equation. Since both sides are now perfect squares, we can set them equal to each other and solve for x and y. This will give us all the possible solutions for the equation.

5. What is the significance of the equation $3x^2+x=4y^2+y$?

This equation is significant because it is an example of a Diophantine equation, which involves finding integer solutions to polynomial equations. It also has a unique property where the left side and right side can be rewritten as perfect squares, making it easier to solve.

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