cooldudeman
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Says to find integer solutions but I only got a non integer. I took a guess about them being perfect squares and not perfect 4ths because if I cant find any solutions if I said they were. What am I doing wrong.

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davros
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(Original post by cooldudeman)
Says to find integer solutions but I only got a non integer. I took a guess about them being perfect squares and not perfect 4ths because if I cant find any solutions if I said they were. What am I doing wrong.

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Check your factorization!
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cooldudeman
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(Original post by davros)
Check your factorization!
? It works out

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BabyMaths
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(Original post by cooldudeman)
? It works out

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Since you're looking for integer solutions I'll post my thoughts.

Solving the quadratic in y and rearranging gives 6y+1=\pm\sqrt{(2x^2)^2+9}.
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cooldudeman
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(Original post by BabyMaths)
Since you're looking for integer solutions I'll post my thoughts.

Solving the quadratic in y and rearranging gives 6y+1=\pm\sqrt{(2x^2)^2+9}.
I kinda forgot to mention that there was a hint saying that I should factorise the RHS.

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firegalley246
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(Original post by cooldudeman)
Says to find integer solutions but I only got a non integer. I took a guess about them being perfect squares and not perfect 4ths because if I cant find any solutions if I said they were. What am I doing wrong.

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I don't think you've done anything wrong - you've managed to prove that there are no integer solutions!

I'm not sure about the reasoning that 3y+2 and 3y-1 are perfect squares though - how did you deduce that?

Here's my take:
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gcd(3y+2, 3y-1) = gcd(3y-1, 3) = 1, which means that both 3y-1 and 3y+2 are fourth powers of integers. But there are no fourth powers that differ by 3; hence there are no integer solutions.
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cooldudeman
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(Original post by firegalley246)
I don't think you've done anything wrong - you've managed to prove that there are no integer solutions!

I'm not sure about the reasoning that 3y+2 and 3y-1 are perfect squares though - how did you deduce that?

Here's my take:
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gcd(3y+2, 3y-1) = gcd(3y-1, 3) = 1, which means that both 3y-1 and 3y+2 are fourth powers of integers. But there are no fourth powers that differ by 3; hence there are no integer solutions.
Yeah I considered the fourth power thing but got no values too. I thpufht that x^4 can be written as (x^2)^2 so taking x^2 as one term being squared therefore the terms on the other side are perfect squares. But I just resorted to doing that because the fourth power thing got me no values.

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davros
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(Original post by cooldudeman)
? It works out

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Sorry, you're right - I was too tired when I looked at it
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firegalley246
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(Original post by davros)
Sorry, you're right - I was too tired when I looked at it
No worries - doing it with squares, I'm sure you would have got most of the marks on the question (if it was in an exam) because it's correct; though doing fourth powers just lets you immediately spot there's no integer solutions, rather than doing squares which gets you only a rational solution.
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DFranklin
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(Original post by cooldudeman)
Says to find integer solutions but I only got a non integer. I took a guess about them being perfect squares and not perfect 4ths because if I cant find any solutions if I said they were. What am I doing wrong.

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It seems your fundamental argument is that since (3y+2)(3y-1) is a perfect square, 3y+2 and 3y-1 must both be perfect squares.

I don't see why this is true; it's certainly not true that if AB is a perfect square, then A and B must be perfect squares.
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cooldudeman
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(Original post by DFranklin)
It seems your fundamental argument is that since (3y+2)(3y-1) is a perfect square, 3y+2 and 3y-1 must both be perfect squares.

I don't see why this is true; it's certainly not true that if AB is a perfect square, then A and B must be perfect squares.
I skipped the part where im meant tp say that 3y+2 and 3y-1 are coprime to each other.

Let h=hcf(3y+2, 3y-1). By a therom h is a factor of (3y+2)-(3y-1)=1. Which means there exists an integer x st 3=xh. So h has to be equal to 1. Since h could be either 3 or 1 but looking at 3y+2 and 3y-1, it cant be 3.
Since 3y+2, 3y-1 are coprime a therom states that they are perfect squares if (3y+2)(3y-1) is a perfect square.
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