- #1

- 8

- 0

a

_{1}- [itex]\sqrt{N}[/itex] b

_{1}= a2 - [itex]\sqrt{N}[/itex]b

_{2}

to be

a

_{1}=a

_{2}and b

_{1}=b

_{2}?

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- Thread starter basil
- Start date

- #1

- 8

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a

to be

a

- #2

- 724

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Matching coefficients?

- #3

- 543

- 1

You can't: Take N = 4, a1=5, b1=1, a2=15, and b2=6

- #4

- 182

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You can't: Take N = 4, a1=5, b1=1, a2=15, and b2=6

In this case, they are only equal because N is a perfect square.

If a + b√N = c + d√N, then

a - c = (d - b)√N

Assuming a, b, c, d are integers, then (a - c) is an integer, thus for equality, (d - b)√N must be an integer. This is only true if a = c and b = d, or if √N is an integer, making N a perfect square.

Last edited:

- #5

- 8

- 0

Hey guys,

I see now...Thanks! Appreciate it.

I see now...Thanks! Appreciate it.

- #6

disregardthat

Science Advisor

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Assuming a, b, c, d are integers, then (a - c) is an integer, thus for equality, (d - b)√N must be an integer. This is only true if a = c and b = d, or if √N is an integer, making N a perfect square.

You skipped a step. sqrt(N) could have been rational, with (d-b) divisible by the denominator. And so this problem is reduced to proving that sqrt(N) cannot be rational for square-free N.

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