Pell's equation 5 and 7

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

D H
Staff Emeritus
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.

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.

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