# Diophanite equation

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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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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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**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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#4

Solving the quadratic in y and rearranging gives .

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

Solving the quadratic in y and rearranging gives .

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

Solving the quadratic in y and rearranging gives .

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**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'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:

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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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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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:

**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:

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.

Show

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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#9

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Sorry, you're right - I was too tired when I looked at it

**davros**)Sorry, you're right - I was too tired when I looked at it

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**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 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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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.

**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.

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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