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Diophanite equation

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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Reply 1
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!
Reply 2
Original post by davros
Check your factorization!


? It works out

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Reply 3
Original post by cooldudeman


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

Solving the quadratic in y and rearranging gives 6y+1=±(2x2)2+96y+1=\pm\sqrt{(2x^2)^2+9}.
Reply 4
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=±(2x2)2+96y+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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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:

Spoiler

Reply 6
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:

Spoiler



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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Reply 7
Original post by cooldudeman


Sorry, you're right - I was too tired when I looked at it :smile:
Original post by davros
Sorry, you're right - I was too tired when I looked at it :smile:


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.
Reply 9
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.
Reply 10
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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