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

[mathslover]

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 ?

 

[bel]

What about y=15?

 

[CRGreathouse]

What about y=15?

For y = 15 there are no solutions, since sqrt(449) isn't an integer... right?

 

[D H]

The first solution pair (x,y) is (1,1). Develop a recursive relationship for the n^th such pair (x_n, y_n) in terms of previous pair(s). The relation you want to prove will fall right out.

 

[Kummer]

Do you the general solution to this Pellian equation? I think that is the way to approach this problem. Followed by induction.

 

[ramsey2879]

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]

$$(1 +\sqrt{2})^n$$ 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