N is a positive integer such that each of 3*N + 1 and 4*N + 1 is a perfect square.
Is N always divisible by 56?
If so, prove it. Otherwise, give a counterexample.
Addressing divisibility by 7.
Approach:
(1) Establish the Pellian and its recursion.
(2) Produce a closed form for N
(3) Show divisibility by 7 of (2) in certain cases
(1)Let 3N+1=a^2, 4n+1=b^2
Then N = (a^2/3-1/3)
Substituting: 4(a^2/3-1/3)+1=b^2
4a^2-3b^2=1, a Pellian with recursive solutions: x(n+1)= 7x(n) + 6y(n), y(n+1) = 8x(n) + 7y(n) (I)
By inspection, {a,b} = {1,1} is a solution, so {13,15} is a solution, as is {181,209} and so on.
(2) However we are interested in N.
Let x(n-1)=X, y(n-1)=Y
Then from (I) (7X+6Y)^2-1 = 3N
49X^2+84XY+36Y^2-1 = 3N
and
(8X+7Y)^2-1 = 4N
64X^2+112XY+49Y^2-1= 4N
Thus 15X^2+28XY+13Y^2=N, equivalently (X+Y)(15X+13Y) = N.(II)
For{X,Y}={1,1} (II) gives N = 56. Naturally, the recursion can be applied repeatedly to produce more complex expressions:
For the next entry we have
(7X+6Y+8X+7Y)(15(7X+6Y)+13(8X+7Y))=N, or
(15X+13Y)(209X+181Y)= N, {x,y}={13,15}
For the next entry we have
(15(7X+6Y)+13(8X+7Y))(209(7X+6Y)+181(8X+7Y))=N, or (209X+181Y)(2911X+2521Y)=N,{x,y}={181,209}
and so on. When X=Y=1, the result is to replicate the numeric values of the current and following values of the sequence as two pairs of sums.
(3)
Write (II) as (X+Y)(14X+14Y+X-Y),
The second term of this expression is thus divisible by 7 iff (X=Y) or (X-Y=7n).
This is surprisingly easy to show, for the congruence system, mod7 of {X,Y} has length of only 4:
Small values of Y,X: sum, {residues mod7}
1-1 = 0 {1,1}
15-13 = 2 {1,6}
209-181 = 28 {6,6}
2911-2521 = 390 {6,1}
40545-35113 = 5432 {1,1}
564719-489061 = 75658 {1,6}
7865521-6811741 = 1053780 {6,6}
109552575-94875313 = 14677262 {6,1}
1525870529-1321442641 = 204427888{1,1}
Thus of each pair of sums, exactly one will be divisible by 7, as was to be shown.
Edited on May 17, 2022, 7:13 am
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Posted by broll
on 2022-05-17 05:47:38 |
10 years later...
