X and y are two natural numbers such that 3x^2+x=4y^2+y. Prove that x-y is the square of a whole number.?

Please show me complete working!!!!!!!!!!!!!!!!!!! GOOD LUCK!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!

Relevance

x and y are natural numbers hence they are countable integers.

They are non negative

given 3x^2+x=4y^2+y

1. 3x^2+x=3y^2+y^2+y

So

2. 3 x^2-3 y^2 =y^2+y-x

from difference between squares

3(x+y)(x-y)=y^2+y-x

so

3. 3(x+y)(x-y)=y^2-(x-y)

divide by (x-y)

4. 3(x+y)=y^2/(x-y)-1

5 3(x+y)+1=y^2/(x-y)

as 3*(x+y)+1 is made up of natural numbers it is natural.

so (x-y) must be a factor of y^2

Now can (x-y)=y i.e x=2y?

No because then from 5

3(x+y)+1=y^2/(x-y) so 3(2*y+y)+1=y^2/(y)

so 9y+1=y which it clearly doesn't so if (x-y) is a factor of y^2 but is not equal to y

so (x-y) is a factor of y^2 but is not equal to y.

this means y cannot be prime. Let y be made up of two prime factors a and b

so y^2=a^2*b^2

(x-y) is a factor of a^2*b^2 x-y cannot equal ab from the reasoning above.

(x-y) cannot be a factor of a*b^2 because it would be a factor of ab again

so (x-y) must be a factor of a^2 or a factor of b^2

We can apply the above reasoning if y has more thna 2 prime factors

In general (x-y) must factor into a^2 b^2. .......etc etc

Back to our two factor y

(x-y) must be a factor of a^2 or b^2

Remembering a and b are prime

considering a

(x-y) is a factor of a or a^2

as a is prime (x-y) can be 1 a or a^2

if (x-y) is 1 then (x-y) is a square proprosition proved

if (x-y)=a^2 then x-y is a square proposition proved.

But what if x-y=a?

Then x=y+a

From 5

3(x+y)+1=y^2/(x-y)

first y=a*b

3(x+y)+1=a^2*b^2/(x-y)

next (x-y)=a

3(x+y)*a+a=a^2*b^2

next y=ab

3(x+ab)*a=a^2*b^2

next x=ab+a

3(ab+a+ab)*a=a^2*b^2

3*a^2(2b+1)=a^2*b^2

3*(2b+1)=b^2

b^2-6b-3=0

b=(6+sqrt(36-12))/2 as sqrt(24) is not a whole number then b is not a whole number but b is a prime number from our original statement so (x-y) cannot=a

We can apply this to b and by extension to any other number of prime factors we can split y into.

so

(x-y)=1 or (x-y)=square of a prime factor of y

Thus (x-y) is a square

QED

I don't think the other proofs work.

Clearly, x>y so write x=y+k and substitue into 3x^2+x = 4y^2+y and then solve for y to get x= 3k +/- sqrt(12k^2+k) Since y is an integer, 12k^2+k is a perfect square. 12k^2+k=(12k+1)*k and 12k+1 and k are relatively prime so both 12k+1 and k must be perfect squares. Therefore k=x-y is a perfect square.

ok, let's try this (it'll be hard with the tiping and everything)

besides, i hope my solution is good enough; i thought it through and it seams ok so far, but i'm not sure :D

3x^2+x=4y^2+y

3x^2-4y^2-y+x=0

3x^2-3y^2+x-y=y^2

3(x^2-y^2)+x-y=y^2

3[(x-y)(x+y)]+x-y=y^2

(x-y)[3(x+y)+1]=y^2

y=sqrt{[3(x+y)+1](x-y)}

y=sqrt[(x-y)]*sqrt{[3(x+y)+1]}

y is a natural number, so i think that would conclude that both (x-y) and [3(x+y)+1] are squares of certain numbers; or at the most it reduces the problem to proving that the two parts, (x-y) and [3(x-y)+1] have no common divisors

hope it helps!

all the best and merry Xmass!

If 3x² + x = 4y² + y

then 3x² - 4y² + x - y = 0

then x - y = 4y² - 3x²

You can prove that (x - y) is the root of a whole number by proving (4y² - 3x²) is the root of a whole number. You can prove this by proving that (4y² - 3x²)² is a whole number. This is evident, knowing that y and x are natural numbers that are being squared. The square of a natural number is always a whole number. So (x - y) is the root of a whole number.