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I'm convinced $5a^2 + 8a + 4 = 4b^2$ can be somehow turned into Pell's equation.

My first steps: Rewrite as $5(a + 4/5)^2 + 4/5 = 4b^2$. Rewrite $B = 2b, A = 5a + 4$ to get $A^2 + 4 = 5B^2$. I'm almost there, I just need to know how to solve this Pell-like equation.

qwr
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  • This may help: Methods of solving pell's equation:http://mathoverflow.net/questions/48251/methods-for-solving-pells-equation – NoChance Oct 18 '15 at 21:55
  • I guess the above equation does not have an integer solution. This is because if you divide both sides by 4, the left hand side can't be factored in the form $(a+k)^2$ for real k. The left hand side has no repeated roots. – NoChance Oct 18 '15 at 22:05
  • @NoChance Smallest solution is $(272, 305)$ – qwr Oct 18 '15 at 22:11
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    Looks like the solutions are $$(a,b)n = \left( \frac15(L{12n+3} - 4), \frac12 F_{12n+3}\right) $$ where $L_k$ and $F_k$ are the $k^{th}$ Lucas and Fibonacci numbers respectively. The first few solutions are $$(272,305),(87840,98209),(28284464,31622993), (9107509824,10182505537)\cdots$$ I'm not sure whether this exhaust all solutions. – achille hui Oct 18 '15 at 22:59
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    Clearly $a$ is even; $a=2k$. Then $(5k+2)^2-5b^2=-1$. You can solve $x^2-5y^2=-1$: Let $(x_1,y_1)$ be the minimal solution of $x^2-5y^2=-1$; the solutions are generated by $(x_1-y_1\sqrt{5})^t=x_n-y_n\sqrt{5}$ with $t$ odd. However, $x\equiv {2,3}\pmod{5}$, not necessarily $x\equiv 2\pmod{5}$, so this won't fully solve the problem. – user236182 Oct 18 '15 at 23:02
  • @user236182 The solutions essentially match. Do you mind posting this as an answer so I can accept it? – qwr Oct 19 '15 at 23:51
  • @achillehui Those are the correct solutions, also given in https://oeis.org/A007805 – qwr Oct 20 '15 at 00:24

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user236182's answer: $a$ must be even, $a = 2k$. Rewrite as $5(4k^2) + 16k +4 = 4b^2$, then multiply by $5/4$ to get $25k^2 + 20k + 5 = 5 b^2 \iff (5k + 2)^2 - 5b^2 = -1$. This is Pell's equation in the form $x^2 - 5y^2 = -1$. This is sufficient for all positive integer solutions.

qwr
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