How does ending in 5 impact the divisibility of x by 7 in Pell's equation?

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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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What about y=15?
 
bel said:
What about y=15?

For y = 15 there are no solutions, since sqrt(449) isn't an integer... right?
 
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.
 
Do you the general solution to this Pellian equation? I think that is the way to approach this problem. Followed by induction.
 
Kummer said:
Do you the general solution to this Pellian equation? I think that is the way to approach this problem. Followed by induction.
(1 +\sqrt{2})^n gives X_{n}\sqrt{2} + Y_{n} where X_{n} and Y_{n} for odd n 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 (X\sqrt{2} + Y)*(1+\sqrt{2})^2 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.
 
ramsey2879 said:
(1 +\sqrt{2})^n gives X_{n}\sqrt{2} + Y_{n} where X_{n} and Y_{n} for odd n 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 (X\sqrt{2} + Y)*(1+\sqrt{2})^2 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 (1+\sqrt{2})^2 = 3+2\sqrt{2} and since (1+\sqrt{2})^3 = 7+5\sqrt{2} then
X_{n} = 0 \mod 5 implies that n = 0 \mod 3
but X_{n} is only a solution for odd n so we calculate (1 + \sqrt{2})^6 = 99 + 70\sqrt{2} so the next solution after n = 3 (x=5, y = 7) is
X_{n+6} = 70*Y_{n} + 99*X_{n} \| Y_{n+6} = 70*2*X_{n} + 99*Y_{n}
From inspection of the above formula it can be seen that
X_{n} = 0 \mod 5 if and only if X_{n+6} = 0 \mod 5 and
Y_{n}=0 \mod 7 if and only if Y_{n+6} = 0 \mod 7
So our proof is complete
 
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