# 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.?

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### 5 Answers

- 1 decade agoFavorite Answer
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

- a_math_guyLv 51 decade ago
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.

- 1 decade ago
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!

- 1 decade ago
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.

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- RosemaryLv 44 years ago
If x and y have opposite signs, then on the right hand side you will be working out the difference between the numbers rather than their sum, so they must have the same signs in order for this identity to work.