Prove that $3(p^2)+61$ is a perfect square only for $p = 1$

Question

It is too be proved that $3(p^2)+61$ is a perfect square only for $$p=1 ; p=N$$ The question arises from a problem in arithmatic progression with a certain soft constraint of natural numbers. At the end a quadratic is formed and since our constraint is of natural number The discriminants should be a perfect square. If we take $p= 1$ we are able to solve the question. Our teacher left it to us to prove $p=1$ is the only solution. I have not proceeded due to lack of ideas.

Answer 1

$q^2-3p^2=61$ is a Pell-like equation.

(The classic Pell equation would have $1$ instead of $61.$)

I believe there are infinitely many solutions to it.

Here are some: $p=1, 6, 10, 25, $ and $39.$

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Addendum (as suggested by trancelocation in a comment to Will Jagy's answer):

$(2+\sqrt3)(2-\sqrt3)=1$ so $q^2-3p^2=61\implies(q+\sqrt3p)(q-\sqrt3p)=61\implies$

$ (q+\sqrt3p)(2+\sqrt3)(q-\sqrt3p)(2-\sqrt3)=61$

$=[(2q+3p)+\sqrt3(q+2p)][(2q+3p)-\sqrt3(q+2p)],$

which means $(2q+3p)^2-3(q+2p)^2=61$,

so for any solution $(p,q)$ to $q^2-3p^2=61$ [including for example ($p,q)=(1,8)],$

there is a solution with larger integers $(2q+3p, q+2p)$.

Answer 2

Given any solution with positive integers $q^2 - 3 p^2 = 61,$ we get an infinite string of (larger) positive solutions by $$ (q,p) \mapsto (2q+3p, \; q+2p) $$

For example, $$ (8,1) \mapsto (19, \; 10) \mapsto (68, \; 39) \; ... $$

$$ (13,6) \mapsto (44, \; 25) \mapsto (163, \; 94) \; ... $$

By Cayley-Hamilton or direct calculation, these can be split into four one-variable recurrences, under the rule $$ x_{n+2} = 4 x_{n+1} - x_n \; .$$

$$ 8, 19, 68, 253, 944, 3523, 13148, 49069, 183128, 683443, 2550644,... $$ $$ 1, 10, 39, 146, 545, 2034, 7591, 28330, 105729, 394586, 1472615, ... $$ $$ 13, 44, 163, 608, 2269, 8468, 31603, 117944, 440173, 1642748, 6130819, ... $$ $$ 6, 25, 94, 351, 1310, 4889, 18246, 68095, 254134, 948441, 3539630,... $$

jagy@phobeusjunior:~$ ./Pell_Target_Fundamental
  Automorphism matrix:  
    2   3
    1   2
  Automorphism backwards:  
    2   -3
    -1   2

  2^2 - 3 1^2 = 1

 w^2 - 3 v^2 = 61 =  61

Wed Sep  4 18:54:19 PDT 2019

w:  8  v:  1  SEED   KEEP +- 
w:  13  v:  6  SEED   BACK ONE STEP  8 ,  -1
w:  19  v:  10
w:  44  v:  25
w:  68  v:  39
w:  163  v:  94
w:  253  v:  146
w:  608  v:  351
w:  944  v:  545
w:  2269  v:  1310
w:  3523  v:  2034
w:  8468  v:  4889
w:  13148  v:  7591
w:  31603  v:  18246
w:  49069  v:  28330
w:  117944  v:  68095
w:  183128  v:  105729
w:  440173  v:  254134
w:  683443  v:  394586
w:  1642748  v:  948441
w:  2550644  v:  1472615
w:  6130819  v:  3539630
w:  9519133  v:  5495874
w:  22880528  v:  13210079
w:  35525888  v:  20510881
w:  85391293  v:  49300686

Wed Sep  4 18:56:21 PDT 2019

 w^2 - 3 v^2 = 61 =  61

jagy@phobeusjunior:~$