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Trying to prove pells equation doesn't work with square numbers watch

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    Hello, I am trying to prove that for pells equation "x^2 - ny^2 = 1" when n is a perfect square the equation has no solutions. Given that all x, y and n are Natural numbers not including zero.
    So far I have got this:
    x^2 - m^2y^2 = 1
    x^2 - (my)^2 = 1
    (x + my) (x - my) = 1 (difference of two squares)
    (x + my) = \frac{1}{(x - my)}

    So from here am I right in saying that the left hand side is going to be 2 or higher and the right surely must be 1 or less? I feel like I'm missing a final step or possibly need to word my statement better, any ideas?
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    (Original post by iAustinMark)
    Hello, I am trying to prove that for pells equation "x^2 - ny^2 = 1" when n is a perfect square the equation has no solutions. Given that all x, y and n are Natural numbers not including zero.
    So far I have got this:
    x^2 - m^2y^2 = 1
    x^2 - (my)^2 = 1
    (x + my) (x - my) = 1 (difference of two squares)
    We have two cases: x+my = x-my = 1 or x+my =  x -my = -1.
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    (Original post by iAustinMark)
    Hello, I am trying to prove that for pells equation "x^2 - ny^2 = 1" when n is a perfect square the equation has no solutions. Given that all x, y and n are Natural numbers not including zero.
    So far I have got this:
    x^2 - m^2y^2 = 1
    x^2 - (my)^2 = 1
    (x + my) (x - my) = 1 (difference of two squares)
    (x + my) = \frac{1}{(x - my)}

    So from here am I right in saying that the left hand side is going to be 2 or higher and the right surely must be 1 or less? I feel like I'm missing a final step or possibly need to word my statement better, any ideas?
    You are home and dry on the second to last line: you have the product of two integers equal to one. What are the possible solutions?
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    (Original post by Gregorius)
    You are home and dry on the second to last line: you have the product of two integers equal to one. What are the possible solutions?
    So (x + my) and (x - my) would have to both equal plus or minus 1? That not being possible if all x,y and m are integers, is that proof enough in itself?
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    (Original post by iAustinMark)
    So (x + my) and (x - my) would have to both equal plus or minus 1? That not being possible if all x,y and m are integers, is that proof enough in itself?
    You now have two sets of equations (one each for +1 , -1) in two unknowns. Solve them. x+my=1 and x-my=1 implies x=? and y=?
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    (Original post by Gregorius)
    You now have two sets of equations (one each for +1 , -1) in two unknowns. Solve them. x+my=1 and x-my=1 implies x=? and y=?
    I suppose it would imply x=1 and y=0 but that's not possible because y can't be 0
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    (Original post by iAustinMark)
    I suppose it would imply x=1 and y=0 but that's not possible because y can't be 0
    Yes.
 
 
 
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