For $5b^2+4=n^2$, prove that $b=\frac{1}{5}(\lambda_2-\lambda_1)(\lambda_2^k-\lambda_1^k)$ where $\lambda_1, \lambda_2$ are roots of $x^2-3x+1=0$.

Question

If $b\in \mathbb{Z}_{>0}$ and $5b^2+4$ is a perfect square, then $b$ can be written in the form $b=\frac{1}{5}(\lambda_2-\lambda_1)(\lambda_2^k-\lambda_1^k)$, where $\lambda_1, \lambda_2$ are the two distinct solutions of the equation $x^2-3x+1=0$.

I have computed that $b=1,3,8, 21$ satisfies the condition and their $k$s are respectively $1,2,3,4$

Answer 1

Lemma $1$: Suppose $5y^2+4=x^2$ for some integer $x,y>0$. Then $5(\frac{3y-x}2)^2+4=(\frac{3x-5y}2)^2$, where $\frac{3y-x}2$, $\frac{3x-5y}2$ are integers, $0\le\frac{3y-x}2<y$, and $0<\frac{3x-5y}2<x$, where the equality $0=\frac{3y-x}2$ holds only when $(x,y)=(3,1)$.
Proof: $5(\frac{3y-x}2)^2-(\frac{3x-5y}2)^2=\frac{5(9y^2-6xy+x^2)-(9x^2-30xy+25y^2)}4=5y^2-x^2=-4.$
$5y^2+4=x^2$ implies $x$ and $y$ are both odd or both even. Hence, $\frac{3y-x}2$ and $\frac{3x-5y}2$ are integers.

• $0\le\frac{3y-x}2\impliedby x^2\le9y^2\impliedby 5y^2+4\le9y^2\impliedby 1\le y^2$
• $\frac{3y-x}2<y\impliedby x>y$
• $0<\frac{3x-5y}2\impliedby 9x^2>25y^2\impliedby 9(5y^2+4)>25y^2\impliedby 20y^2+36>0$
• $\frac{3x-5y}2<x\impliedby x<5y\impliedby x^2<25y^2\impliedby 5y^2+4<25y^2\impliedby 4 < 20 y^2$

Lemma $2$: Suppose $5y^2+4=x^2$ for some integer $x,y\ge0$. Then $5(\frac{x+3y}2)^2+4=(\frac{3x+5y}2)^2$, where $\frac{x+3y}2$, $\frac{3x+5y}2$ are integers, $y<\frac{x+3y}2$ and $x<\frac{3x+5y}2$.
Proof. Similar to the proof above.

Lemma $3$: The transformations $(x,y)\to(\frac{3x-5y}2, \frac{3y-x}2)$ and $(x,y)\to(\frac{3x+5y}2,\frac{x+3y}2)$ are inverse to each other.
Proof. It is straightforward.

Theorem. All nonnegative integer solutions of $5y^2+4=x^2$ are given by $$\begin{aligned} x&=\lambda_1^i+\lambda_2^i\\ y&=\frac{\lambda_1^i-\lambda_2^i}{\sqrt 5}=\frac{1}{5}(\lambda_2-\lambda_1)(\lambda_2^i-\lambda_1^i) \end{aligned}$$ where $i\in\Bbb N$, $\lambda_1=\frac{3+\sqrt5}2, \lambda_2=\frac{3-\sqrt5}2$ are the two roots of $x^2-3x+1=0$.
Proof. The lemmas implies that starting from $(x_0,y_0)=(2,0)$ and computing $(x_{i+1}, y_{i+1})=(\frac{3x_i+5y_i}2,\frac{x_i+3y_i}2)$ inductively for all $i\ge0$, we will obtain all nonnegative pairs of integers $(x,y)$ with $5y^2+4=x^2$.

• $(2,0)=(\lambda_1^0+\lambda_2^0, (\lambda_1^0-\lambda_2^0)/\sqrt5)$
• Suppose $(x_i, y_i)=(\lambda_1^i+\lambda_2^i,\frac{\lambda_1^i-\lambda_2^i}{\sqrt 5})$. Then $$x_{i+1}=\frac{3(\lambda_1^i+\lambda_2^i)+5\frac{\lambda_1^i-\lambda_2^i}{\sqrt5}}2=\frac{3+\sqrt5}2\lambda_1^i+\frac{3-\sqrt5}2\lambda_2^i=\lambda_1^{i+1}+\lambda_2^{i+1}$$ $$y_{i+1}=\frac{(\lambda_1^i+ \lambda_2^i)+3\frac{\lambda_1^i-\lambda_2^i}{\sqrt5}}2=\frac{3+\sqrt5}{2\sqrt5}\lambda_1^i-\frac{3-\sqrt5}{2\sqrt5}\lambda_2^i=\frac{\lambda_1^{i+1}-\lambda_2^{i+1}}{\sqrt5}$$

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For readers who love to see the actual integers, here are the initial 20 pairs of nonnegative integer solutions.

$(x,y)$ with $5y^2+4=x^2$.
$(2, 0)$
$(3, 1)$
$(7, 3)$
$(18, 8)$
$(47, 21)$
$(123, 55)$
$(322, 144)$
$(843, 377)$
$(2207, 987)$
$(5778, 2584)$
$(15127, 6765)$
$(39603, 17711)$
$(103682, 46368)$
$(271443, 121393)$
$(710647, 317811)$
$(1860498, 832040)$
$(4870847, 2178309)$
$(12752043, 5702887)$
$(33385282, 14930352)$
$(87403803, 39088169)$

Answer 2

COMMENT.-Your equation $5b^2+4=k^2$ is equivalent to $$\left(\dfrac k2\right)^2-5\left(\dfrac b2\right)^2=1$$ and the Pell-Fermat equation $$x^2-5y^2=1$$ has the infinite set of integer solutions where $n$ is natural integer. $$x=\frac{(9+4\sqrt5)^n+(9-4\sqrt5)^n}{2}\\y=\frac{(9+4\sqrt5)^n-(9-4\sqrt5)^n}{2\sqrt5}$$ so you have integer solution for $k,b$ $$k=(9+4\sqrt5)^n+(9-4\sqrt5)^n\\b=\frac{(9+4\sqrt5)^n-(9-4\sqrt5)^n}{\sqrt5}$$ With these solutions you can verify if it is true what you say.