Pell's equation 5 and 7

  • Thread starter mathslover
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  • #1
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guys,

From the solutions of the Pell's equation x*x-2*y*y=-1,
how can we prove that whenever y ends in digit 5, then 7 | x ?

-Mathslover

Perhaps I should clarify a bit,x*x-2*y*y=-1 has solution
x=1, 7, 41, 239, 1393, 8119, 47321, 275807,.....
y=1, 5, 29, 169, 985, 5741, 33461, 167305,...
and the general solution is (xn+yn*sqrt(2))=(1+sqrt(2))^(2*n+1)
Apart from induction,how can we prove whenever 5|y then 7|x ?
 
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Answers and Replies

  • #2
bel
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What about y=15?
 
  • #3
CRGreathouse
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What about y=15?
For y = 15 there are no solutions, since sqrt(449) isn't an integer... right?
 
  • #4
D H
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The first solution pair (x,y) is (1,1). Develop a recursive relationship for the nth such pair (xn, yn) in terms of previous pair(s). The relation you want to prove will fall right out.
 
  • #5
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Do you the general solution to this Pellian equation? I think that is the way to approach this problem. Followed by induction.
 
  • #6
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Do you the general solution to this Pellian equation? I think that is the way to approach this problem. Followed by induction.
[tex](1 +\sqrt{2})^n[/tex] gives [tex]X_{n}\sqrt{2} + Y_{n}[/tex] where [tex]X_{n}[/tex] and [tex]Y_{n}[/tex] for odd [tex]n[/tex] are solutions.

thus the first 3 solutions pairs X,Y are
1,1
5,7
29,41

from this one might notice that 5 = 3*1+2*1 and 29 = 3*5+2*7 and also notice that
7 = 4*1 + 3*1 and that 41=4*5+3*7.
Or simply multiply [tex](X\sqrt{2} + Y)*(1+\sqrt{2})^2[/tex] and put back into the same X,Y form to find the new X and Y.

So prove that if X and Y are a solution pair then (3X+2Y) and (4X+3Y) are the next solution pair then consider X mod 5 and Y mod 7 and you can follow the rest by induction.
 
  • #7
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[tex](1 +\sqrt{2})^n[/tex] gives [tex]X_{n}\sqrt{2} + Y_{n}[/tex] where [tex]X_{n}[/tex] and [tex]Y_{n}[/tex] for odd [tex]n[/tex] are solutions.

thus the first 3 solutions pairs X,Y are
1,1
5,7
29,41

from this one might notice that 5 = 3*1+2*1 and 29 = 3*5+2*7 and also notice that
7 = 4*1 + 3*1 and that 41=4*5+3*7.
Or simply multiply [tex](X\sqrt{2} + Y)*(1+\sqrt{2})^2[/tex] and put back into the same X,Y form to find the new X and Y.

So prove that if X and Y are a solution pair then (3X+2Y) and (4X+3Y) are the next solution pair then consider X mod 5 and Y mod 7 and you can follow the rest by induction.
Since [tex](1+\sqrt{2})^2 = 3+2\sqrt{2}[/tex] and since [tex](1+\sqrt{2})^3 = 7+5\sqrt{2}[/tex] then
[tex]X_{n} = 0 \mod 5[/tex] implies that [tex] n = 0 \mod 3 [/tex]
but [tex]X_{n}[/tex] is only a solution for odd n so we calculate [tex](1 + \sqrt{2})^6 = 99 + 70\sqrt{2}[/tex] so the next solution after n = 3 (x=5, y = 7) is
[tex]X_{n+6} = 70*Y_{n} + 99*X_{n} \| Y_{n+6} = 70*2*X_{n} + 99*Y_{n}[/tex]
From inspection of the above formula it can be seen that
[tex]X_{n} = 0 \mod 5[/tex] if and only if [tex] X_{n+6} = 0 \mod 5[/tex] and
[tex]Y_{n}=0 \mod 7[/tex] if and only if [tex] Y_{n+6} = 0 \mod 7[/tex]
So our proof is complete
 

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