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Find all natural numbers $n$ such that $21n^2-20$ is a perfect square.

I have got the following solutions via programming:

$n=1,2,3,9,14,43,67$ but how can I find these manually? How can I guarantee there are no more solutions?

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    $\begingroup$ Look at this oeis.org/… $\endgroup$ – Vincent May 5 '16 at 11:46
  • $\begingroup$ but is the manual method given? $\endgroup$ – Legend Killer May 5 '16 at 11:48
  • $\begingroup$ Did you read the linked site? $\endgroup$ – flawr May 5 '16 at 11:49
  • $\begingroup$ No, but maybe by twisting your equations, you might find theirs. $\endgroup$ – Vincent May 5 '16 at 11:49
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    $\begingroup$ are talking about the recurrence $a_n=5a_{n-2}+a_{n-4}$? $\endgroup$ – Legend Killer May 5 '16 at 11:53
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Now that I see we can drop the discriminant, this is better, and very similar to what Dietrich wrote:

IF $$ x^2 + xy - 5 y^2 = -5, $$ then $$ 4 x^2 + 4xy - 20 y^2 = -20, $$ $$ (2x+y)^2 - 21 y^2 = -20. $$ Notice that this gives everything, because if $u^2 - 21 v^2 = -20,$ then $u \equiv v \pmod 2.$

So, to get fewer orbits with the nicer recursion: we get $$ y_{k+2} = 5 y_{k+1} - y_k $$ in two threads $$ 1, 3, 14, 67, 321, 1538, 7369,... $$ $$ 2, 9, 43, 206, 987, 4729, 22658,... $$ A really careful proof of this is a Conway Topograph, which is easy enough for this problem that I will probably draw it and post it here.

enter image description here

jagy@phobeusjunior:~$  ./Pell_Target_Fundamental_A
2  5
1  3


  5^2 - 21 1^2 = 4

1 x^2 + 1 x y -5 y^2 = -5

Thu May  5 14:14:32 PDT 2016

x:  0  y:  1 ratio: 0  seed 
x:  3  y:  2 ratio: 1.5  seed 
x:  5  y:  3 ratio: 1.666666666666667  seed 
x:  16  y:  9 ratio: 1.777777777777778
x:  25  y:  14 ratio: 1.785714285714286
x:  77  y:  43 ratio: 1.790697674418605
x:  120  y:  67 ratio: 1.791044776119403
x:  369  y:  206 ratio: 1.79126213592233
x:  575  y:  321 ratio: 1.791277258566978
x:  1768  y:  987 ratio: 1.79128672745694
x:  2755  y:  1538 ratio: 1.791287386215865
x:  8471  y:  4729 ratio: 1.79128779868894
x:  13200  y:  7369 ratio: 1.791287827384991
x:  40587  y:  22658 ratio: 1.791287845352635
x:  63245  y:  35307 ratio: 1.791287846602656
x:  194464  y:  108561 ratio: 1.791287847385341
x:  303025  y:  169166 ratio: 1.791287847439793
x:  931733  y:  520147 ratio: 1.791287847473887
x:  1451880  y:  810523 ratio: 1.791287847476259
x:  4464201  y:  2492174 ratio: 1.791287847477744
x:  6956375  y:  3883449 ratio: 1.791287847477848

Thu May  5 14:14:52 PDT 2016
2  5
1  3
Inverse of given automorphism of quadratic form:
3  -5
-1  2
jagy@phobeusjunior:~$
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Note that $$ x^2−5xy+y^2+5=0 $$ is equivalent to $$ 21x^2-20=(2y-5x)^2. $$ Hence all solutions of the first equation are also solutions of $21n^2-20=m^2$. This explains that the solutions given at OEIS (see the above comment) are also solutions here.

In general, it is well known how to solve the quadratic equations $ax^2+bxy+cy^2=k$ over the integers. In particular, we can solve the generalised Pell's equation $$ 21x^2-y^2=20. $$

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From theory of Pell equation $$21\Bigl(\frac{(55+12\sqrt{21})^n(1+\sqrt{21})-(55-12\sqrt{21})^n(1-\sqrt{21})}{2\sqrt{21}}\Bigr)^2-20=\Bigl(\frac{(55+12\sqrt{21})^n(1+\sqrt{21})+(55-12\sqrt{21})^n(1-\sqrt{21})}{2}\Bigr)^2.$$

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