Now considering the expression of $\Lambda$ in (7.30) and taking into account
(7.32) we have

$\left|\frac{1}{\pi}\mathrm{Arg}\left(\frac{1+\xi i}{1-\xi
i}\right)-\frac{k}{v}\right|<\frac{4.81}{\pi v(1.75\sqrt{a})^{v}}.$

We put $\gamma=\pi^{-1}\mathrm{Arg}((1+\xi i)/(1-\xi i))$. Then,

(7.36) $\left|\gamma-\frac{k}{v}\right|<\frac{4.81}{\pi
v(1.75\sqrt{a})^{v}}<\frac{1}{2v^{2}}$

which shows that $k/v$ is a convergent to the continued fraction expansion of
$\gamma$, with $v<V=25,000$.

The values of $a$ and $v$ are now small enough that we can use the continued-
fraction algorithm to complete the proof. We denote by $p_{i}/q_{i}$
($i=0,1,2,\ldots$) the convergents of this expansion.212121Obviously, all
these depend on $a$. Let $h$ be the least $i$ such that $q_{i}\geq V$. Then,
$k/v=p_{j}/q_{j}$222222Note that this equality does not necessarily mean that
$k=p_{j}$ and $v=q_{j}$; in general $v\geq q_{j}$ with $0\leq j\leq h-1$. For
each such $k/v$ we calculate both sides of (7.36).

We did so for all possible values of $750\leq a\leq 915,000,000$ and all
admissible convergents. We found no examples where (7.36) is satisfied and
this completes the proof when $a\geq 750$.

We used pari/gp Version 2.12.0 [10] to perform this search. It took 15 hours
and 4 minutes on a laptop with an Intel i7-3630QM processor and 8gb of RAM.

It remains to check the values $4\leq a<750$. For these values, we go back, to
a few lines below (7.32) and follow the same steps which led us to (7.34) but,
in order to obtain a lower bound for $|\Lambda|$, instead of using Theorem 2
of [16], we apply Théorème 3 of [7] whose application imposes no condition on
$a$ at the cost of obtaining a larger upper bound for $v$, namely, $v<2.8\cdot
10^{5}$. This is not a problem at all. We repeat the continued fraction
procedure which we applied immediately below (7.36) for each value
$a=4,\ldots,749$, checking all convergents $k/v$ with $v<2.8\cdot 10^{5}$.
Except for the case $a=7$, no examples where (7.36) is satisfied. When $a=7$
the convergent $p_{2}/q_{2}=3/4$ gives $k/v=3/4$ which satisfies (7.36) but
then we turn to (7.30) and see that this is not satisfied. This completes the
proof for all values of $a$. ∎

###  7.3 Impossibility of the equation in (IIIc) of Section 6

######  Theorem 7.10.

If $a=c^{2}+1$ with $c\geq 2$, then the equation $w_{n}=x^{2}$ with odd $n\geq
3$, is impossible.

###### Proof.

From the relations $x^{2}=w_{n}\geq w_{3}$ and $a\geq 5$ we have $x>9$. We put
$n=2m+1\geq 3$. By Proposition 5.7, we have
$x\xi+\alpha^{m+1}=\alpha^{u}(c+\xi)^{v}$ with $x,u,v\in{\mathbb{Z}}$. Along
with this equation we have also its conjugate equations (cf. just above Lemma
3.18)

$\displaystyle-x\xi+\alpha^{m+1}$ $\displaystyle=\alpha^{u}(c-\xi)^{v}$ (7.37)
$\displaystyle x\phi i+\beta^{m+1}$ $\displaystyle=\beta^{u}(c+\phi i)^{v}$
$\displaystyle-x\phi i+\beta^{m+1}$ $\displaystyle=\beta^{u}(c-\phi i)^{v}.$

We claim that $v>0$. Indeed, we have
$2x\xi=a^{u}\left((c+\xi)^{v}-(c-\xi)^{v}\right)$ and $c+\xi>c-\xi$. If $v\leq
0$, then the quantity in the big parentheses would be $\leq 0$, which
contradicts the fact that both $x$ and $\xi$ are positive.

From the last two equations in (7.3) we have

(7.38) $\left(\frac{c+\phi i}{c-\phi i}\right)^{\\!\\!v}=\frac{x\phi
i+\beta^{m+1}}{-x\phi i+\beta^{m+1}}=-1+\frac{2\beta^{m+1}}{-x\phi
i+\beta^{m+1}}.$

From (5.9), we have $(\alpha-1)x^{2}=\alpha^{2m+2}-\beta^{2m+1}$ and
considering its conjugate relation we have
$(\beta-1)x^{2}=\beta^{2m+2}-\alpha^{2m+1}$ from which it follows that
$(1-\beta)x^{2}+\beta^{2m+2}=\alpha^{2m+1}$. Then, $\left|-x\phi
i+\beta^{m+1}\right|^{2}=x^{2}\phi^{2}+\beta^{2m+2}=(1-\beta)x^{2}+\beta^{2m+2}=\alpha^{2m+1}$
and consequently

(7.39) $\left|\frac{2\beta^{m+1}}{-x\phi
i+\beta^{m+1}}\right|=\frac{2\beta^{m+1}}{\alpha^{m+\frac{1}{2}}}=2\beta^{2m+\frac{3}{2}}<1$

Now we temporarily put $z=2\beta^{m+1}/(-x\phi i+\beta^{m+1})$, so that
$|z|<2\beta^{2m+\frac{3}{2}}$ and $|z|^{2}<2\beta^{2m+\frac{3}{2}}|z|$, and
define

$\displaystyle\Lambda$ $\displaystyle:=\mathrm{Log}\left(-((c+\phi i)/(c-\phi
i))^{v}\right)=\mathrm{Log}\left(1-\frac{2\beta^{m+1}}{-x\phi
i+\beta^{m+1}}\right)$
$\displaystyle=\mathrm{Log}(1-z)=-\sum_{n=1}^{\infty}\frac{z^{n}}{n},$

the second equality in the first line holding by (7.38). Then
$|\Lambda|<|z|+|z|^{2}<(1+2\beta^{2m+\frac{3}{2}})|z|\leq(1+2\beta^{7/2})|z|<1.06889|z|<\mbox{(by
\eqref{eq 2 solving the xi case})}\>2.138\beta^{2m+\frac{3}{2}}$. Thus,

(7.40) $\Lambda=\mathrm{Log}\left(-((c+\phi i)/(c-\phi
i))^{v}\right)\quad\mbox{and}\quad|\Lambda|<2.138\beta^{2m+\frac{3}{2}}.$

In general, $\mathrm{Arg}(\zeta)-\mathrm{Arg}(-\zeta)=\pm\pi$ for $\zeta\neq
0$. Therefore,

$\displaystyle\Lambda$ $\displaystyle=\mathrm{Log}\left(-\left(\frac{c+\phi
i}{c-\phi
i}\right)^{\\!\\!v}\right)=i\cdot\mathrm{Arg}\left(\left(\frac{c+\phi
i}{c-\phi i}\right)^{\\!\\!v}\right)+\nu\pi i\qquad\mbox{($\nu=\pm 1$)}$
$\displaystyle=\mathrm{Log}\left(\left(\frac{c+\phi i}{c-\phi
i}\right)^{\\!\\!v}\right)+\nu\pi i.$

In view of the general formula
$\mathrm{Log}(\zeta^{n})=n\cdot\mathrm{Log}\,\zeta+2\kappa\pi i$ we have
$\displaystyle{\mathrm{Log}\left(\left(\frac{c+\phi i}{c-\phi
i}\right)^{\\!\\!v}\right)=v\cdot\mathrm{Log}\left(\frac{c+\phi i}{c-\phi
i}\right)+2\kappa\pi i}$. Hence

$-(v+1)\pi<2\kappa\pi=\mathrm{Arg}\left(\left(\frac{c+\phi i}{c-\phi
i}\right)^{v}\right)-v\,\mathrm{Arg}\left(\frac{c+\phi i}{c-\phi
i}\right)<(v+1)\pi.$

Consequently, $2|\kappa|\leq v$. Then, $\Lambda=v\mathrm{Log}((c+\phi
i)/(c-\phi i))+(2\kappa+\nu)\pi i$ and putting $2\kappa+\nu=-k$ we obtain

$\Lambda=v\cdot\mathrm{Log}\left(\frac{c+\phi i}{c-\phi i}\right)-k\pi
i,\quad|k|\leq v+1.$

######  Lemma 7.11.

We have

(7.41) $\log\left|\Lambda\right|<-|v|\log(2c^{2})+\log(4.781).$

###### Proof.

We claim that $k>0$. Note that $|\Lambda|=|v\cdot\mathrm{Arg}((c+\phi
i)/(c-\phi i)-k\pi|$. A computation with Maple shows that
$\mathrm{Arg}((c+\phi i)/(c-\phi i))>2.138\beta^{7/2}$; in particular,
$\mathrm{Arg}((c+\phi i)/(c-\phi i))>0$. If $k\leq 0$, then (since $v>0$)
$|\Lambda|\geq\mathrm{Arg}((c+\phi i)/(c-\phi i))>2.138\beta^{7/2}$. On the
other hand, from the upper bound for $|\Lambda|$ in (7.40),
$|\Lambda|<2.138\beta^{2m+3/2}\leq 2.138\beta^{7/2}$ and we get a
contradiction. Thus, we finally get

(7.42) $\Lambda=v\cdot\mathrm{Log}\left(\frac{c+\phi i}{c-\phi i}\right)-k\pi
i,\quad 0<k\leq v+1,\quad|\Lambda|<2.138\cdot\beta^{2m+\frac{3}{2}}.$

We will use this upper bound for $|\Lambda|$ in order to bound $|\Lambda|$ in
terms of $v$. To accomplish this, we will consider the quantity $\left|(c+\phi
i)(c+\xi)\right|^{-2v}$ and show it is larger than $\beta^{4m+1}/5$. The
desired upper bound for $|\Lambda|$ will then follow from (7.42).

We have

$\displaystyle(c+\phi i)^{v}(c+\xi)^{v}$ $\displaystyle=(x\phi
i+\beta^{m+1})(x\xi+\alpha^{m+1})$ $\displaystyle=(1+\xi\beta^{m+1}x)+(\xi\phi
x^{2}+\phi\alpha^{m+1}x)i.$

Therefore,

$|(c+\phi i)(c+\xi)|^{2v}=(1+\xi\beta^{m+1}x)^{2}+(\xi\phi
x^{2}+\phi\alpha^{m+1}x)^{2},$

and we are looking for an upper bound for the right-hand side. We have
$0<1+\xi\beta^{m+1}x<1+\xi\beta x\leq(1+\xi\beta)x<\xi\phi x\leq\xi\phi
x^{2}$. From the very beginning of the proof we have
$-x\xi+\alpha^{m+1}=\alpha^{u}(c-\xi)^{v}>0$ and multiplying this by $\phi x$
we see that $-\xi\phi x^{2}+\phi\alpha^{m+1}x>0$. Thus,
$\phi\alpha^{m+1}x>\xi\phi x^{2}>1+\xi\beta^{m+1}x$. Hence the right-hand side
of the above displayed equality is
$<(\phi\alpha^{m+1}x)^{2}+(2\phi\alpha^{m+1}x)^{2}=5(\phi\alpha^{m+1}x)^{2}$
hence, $|(c+\phi i)(c+\xi)|^{v}<\sqrt{5}\phi\alpha^{m+1}x$ and consequently,

(7.43) $|(c+\phi
i)(c+\xi)|^{-2v}>\frac{1}{5\phi^{2}}\cdot\frac{\beta^{2m+2}}{x^{2}}=\frac{\beta^{2m+2}}{5(1-\beta)}\cdot\frac{1}{x^{2}}$

From (5.9), we see that $(\alpha-1)x^{2}=\alpha^{2m+2}-\beta^{2m+1}$.
Therefore

$\displaystyle\frac{1}{x^{2}}$
$\displaystyle=\frac{\alpha-1}{\alpha^{2m+2}-\beta^{2m+1}}=\frac{1-\beta}{\beta(\alpha^{2m+2}-\beta^{2m+1})}=\frac{1-\beta}{\beta^{-2m-1}-\beta^{2m+2}}$
$\displaystyle=\frac{(1-\beta)\beta^{2m+1}}{1-\beta^{4m+3}}>(1-\beta)\beta^{2m+1}.$

This combined with (7.43) gives $|(c+\phi i)(c+\xi)|^{-2v}>\beta^{4m+3}/5$.
Therefore, $\beta^{2m+3/2}<\sqrt{5}|(c+\phi i)(c+\xi)|^{-v}$ which, combined
with (7.40) implies that

(7.44) $|\Lambda|<2.138\cdot\sqrt{5}\cdot|(c+\phi
i)(c+\xi)|^{-v}<2.138\cdot\sqrt{5}\cdot(2c^{2})^{-v},$

where the last but one inequality is due to the fact that $|(c+\phi
i)(c+\xi)|\geq|\mathrm{Re}((c+\phi i)(c+\xi))|=c^{2}+c\xi>2c^{2}$. The lemma
follows. ∎

Now we will apply Theorem 2 of [16] to the above linear form $\Lambda$.

######  Lemma 7.12.

We have

(7.45) $\log\left|\Lambda\right|>-469\left((\log a+\log
5)/4\right)\log^{2}(|v|+1).$

###### Proof.

In the notation of [16], we (temporarily) set $\gamma=(c+\phi i)/(c-\phi i)$,
$b_{1}=k$, $b_{2}=v$ and $d=4$.

A symbolic computation shows that the minimal polynomial of $\gamma$ is
$X^{4}-4(c^{4}+c^{2}-1)X^{3}+2(4c^{4}-4c^{2}+3)X^{2}-4(c^{4}+c^{2}-1)X+1$
whose roots are $\gamma$, $\gamma^{-1}$ and $(c\pm\xi)/(c\mp\xi)$. It follows
easily that the absolute logarithmic height of $\gamma$ is
$\frac{1}{4}\log((c+\xi)/(c-\xi))$. We have
$0<(c+\xi)/(c-\xi)=\alpha(c+\xi)^{2}<(c^{2}+1)(2c)^{2}<5c^{4}$. Therefore,
$\mathrm{h}(\gamma)<\log c+\frac{1}{4}\log 5$.

Note that for $c\geq 6$, we have $\mathrm{h}(\gamma)>2$ and so for such $c$,
the hypothesis that $d\,\mathrm{h}(\gamma)>8$, necessary for the application
of Theorem 2 of [16], holds.

Finally, we set
$B=\max\left\\{2,\left|b_{1}\right|,\left|b_{2}\right|\right\\}=\max\\{2,v,k\\}\leq
v+1$, since $v\geq 1$.

In fact, we will assume that $v>24,000$. Then,

$22.36d^{3}(1/2+0.28/\log B+1.8/4/\log B)^{2}<469.$

So, from according to Theorem 2(i) of [16],

$|\Lambda|\geq\exp\left\\{-469\mathrm{h}(\gamma)\log^{2}(B)\right\\},$

and the lemma follows. ∎

Combining this lower bound for $|\Lambda|$ with (7.44), we obtain

$v<\frac{\log 4.8478}{\log(2c^{2})}+\frac{469(\log c+\tfrac{1}{4}\log
5)}{\log(2c^{2})}\log^{2}(v+1)<0.37+241\cdot\log^{2}(v+1).$

The final inequality is impossible for $v\geq 25,000$ and thus we find an
upper bound

(7.46) $v<V:=25,000.$

######  Lemma 7.13.

We have

(7.47) $c<16,000.$

###### Proof.

We write

$\frac{c+\phi i}{c-\phi i}=\frac{z}{\bar{z}},\;\;\mbox{where
$z=1+(\phi/c)i$},\quad 0<\mathrm{Arg}\,z<\pi/4,\;0<\phi/c<1.$

Note that, in view of $0<\mathrm{Arg}\,z<\pi/4$, we have
$\mathrm{Arg}(z^{2})=2\mathrm{Arg}\,z$ and
$\mathrm{Arg}\,\bar{z}=-\mathrm{Arg}\,z$. Therefore,

$\mathrm{Log}(z/\bar{z})=\mathrm{Log}(z/\bar{z})=i\cdot\mathrm{Arg}(z/\bar{z})=i\cdot\mathrm{Arg}(z^{2}/|z|^{2})=i\cdot\mathrm{Arg}(z^{2})\\\
=2i\cdot\mathrm{Arg}\,z$

and

$\mathrm{Log}\,z-\mathrm{Log}\,\bar{z}=i\cdot(\mathrm{Arg}\,z-\mathrm{Arg}\,\bar{z})=i\cdot(\mathrm{Arg}\,z-(-\mathrm{Arg}\,z))=2i\cdot\mathrm{Arg}\,z.$

Consequently, $\mathrm{Log}(z/\bar{z})=\mathrm{Log}\,z-\mathrm{Log}\,\bar{z}$
and

$\mathrm{Log}\left(\frac{c+\phi i}{c-\phi
i}\right)=\mathrm{Log}(z/\bar{z})=\mathrm{Log}\,z-\mathrm{Log}\,\bar{z}.$

We also have the series expansions

$\mathrm{Log}\,z=i(\phi/c)+\frac{1}{2}(\phi/c)^{2}-\frac{i}{3}(\phi/c)^{3}-\frac{1}{4}(\phi/c)^{4}+\frac{i}{5}(\phi/c)^{5}+\frac{1}{6}(\phi/c)^{6}-\cdots,$
$\mathrm{Log}\,\bar{z}-i(\phi/c)+\frac{1}{2}(\phi/c)^{2}+\frac{i}{3}(\phi/c)^{3}-\frac{1}{4}(\phi/c)^{4}-\frac{i}{5}(\phi/c)^{5}+\frac{1}{6}(\phi/c)^{6}+\cdots.$

Therefore,

$\Lambda=v\,\mathrm{Log}(z/\bar{z})-k\pi
i=i\cdot\left(2v\left[(\phi/c)-\textstyle{\frac{1}{3}}(\phi/c)^{3}+\textstyle{\frac{1}{5}}(\phi/c)^{5}-\textstyle{\frac{1}{7}}(\phi/c)^{7}+\cdots\right]-k\pi\right).$

Thus

(7.48) $\left|2vf(\phi/c)-k\pi\right|=|\Lambda|<4.781\cdot(2c^{2})^{-v},$

where $f(z)=z-z^{3}/3+z^{5}/5-z^{7}/7+\cdots$. A symbolic computation shows
that

(7.49)
$\frac{1}{c}-\frac{5}{6c^{3}}<f(\phi/c)<\frac{1}{c}-\frac{5}{6c^{3}}+\frac{43}{40c^{5}}$

If $v\geq\pi c/2$, then $c\leq 2v/\pi<(2/\pi)\cdot 25,000<16,000$ in view of
(7.46). If we succeed in showing that the relation $v<\pi c/2$ leads to a
contradiction, then the lemma will follow.

So, now let us assume that $v<\pi c/2$. In (7.48) we distinguish two cases.
(i) If $2vf(\phi/c)\geq k\pi$, then $k\leq(2v/\pi)\cdot
f(\phi/c)<1-5/(6c^{2})+43/(40c^{4})<1$. Hence $k=0$ which contradicts (7.42).
(ii) If $2vf(\phi/c)<k\pi$, then $|\Lambda|=k\pi-2vf(\phi/c)$. We have
$2vf(\phi/c)<\pi cf(\phi/c)<\pi\,\left(1-5/(6c^{2})+43/(40c^{4})\right)$, by
applying (7.49). Therefore,

(7.50) $\displaystyle 4.781(2c^{2})^{-v}$
$\displaystyle>|\Lambda|=k\pi-2vf(\phi/c)$
$\displaystyle>(k-1)\pi+\pi\left(5/(6c^{2})-43/(40c^{4})\right).$

Obviously, this inequality is impossible if $k\geq 2$. By (7.42), we have
$k>0$. Therefore $k=1$. If $v\geq 2$, then (7.50) implies
$4.781/(4c^{4})>\pi\left(5/(6c^{2})-43/(40c^{4})\right)$ and we check that
this is false for $c\geq 2$. If $v=1$, then $|\Lambda|=|\mathrm{Arg}((c+\phi
i)/(c-\phi i))-\pi|$. Since $0<\mathrm{Arg}((c+\phi i)/(c-\phi i))<2/c\leq 1$,
it follows that $|\Lambda|=\pi-\mathrm{Arg}((c+\phi i)/(c-\phi i))>\pi-2/c$
and now (7.50) implies that $4.781/(2c^{2})>\pi-2/c$ which is false. The lemma
follows. ∎

Now, considering the expression of $\Lambda$ in (7.42) and taking into account
(7.44) we have

$\left|\frac{1}{\pi}\mathrm{Arg}\left(\frac{c+\phi i}{c-\phi
i}\right)-\frac{k}{v}\right|<\frac{4.781}{\pi v(2c^{2})^{v}}.$

We put $\gamma=\pi^{-1}\mathrm{Arg}((c+\phi i)/(c-\phi i))$. Then,

(7.51) $\left|\gamma-\frac{k}{v}\right|<\frac{4.781}{\pi
v(2c^{2})^{v}}<\frac{1}{2v^{2}}$

which shows that $k/v$ is a convergent to the continued fraction expansion of
$\gamma$, with $v\leq V=25,000$. The values of $c$ and $v$ are now small
enough that we can use the continued-fraction algorithm to complete the proof.
We denote by $p_{i}/q_{i}$ ($i=0,1,2,\ldots$) the convergents of this
expansion.232323Obviously, all these depend on $c$. Let $h$ be the least $i$
such that $q_{i}\geq V$. Then, $k/v=p_{j}/q_{j}$242424Note that this equality
does not necessarily mean that $k=p_{j}$ and $v=q_{j}$; in general $v\geq
q_{j}$ with $0\leq j\leq h-1$. For each such $k/v$ we calculate both sides of
(7.51).

We did so for all possible values of $6\leq c\leq 16,000$ and all admissible
convergents. We found no examples where (7.51) is satisfied and this completes
the proof when $c\geq 6$.

We used PARI/GP Version 2.12.0 [10] to perform this search. It took 1.5
seconds on a laptop with an Intel i7-3630QM processor and 8gb of RAM.

It remains to check the values $c=2,3,4,5$. For these values, we go back to a
few lines below (7.44) and follow the same steps which led us to (7.46) but,
in order to obtain a lower bound for $|\Lambda|$, instead of using Theorem 2
of [16], we apply Théorème 3 of [7] whose application imposes no condition on
$c$ at the cost of obtaining a larger upper bound for $v$, namely, $v<2\cdot
10^{5}$. This is not a problem at all. We repeat the continued fraction
procedure which we applied immediately below (7.51) for each value
$c=2,3,4,5$, checking all convergents $k/v$ with $v<2\cdot 10^{5}$. Again we
found no examples where (7.51) is satisfied and this completes the proof for
all values of $c$.

Using the same software and hardware as above, this calculation took under 1
second. ∎

##  8 Interdependence of the results of this paper

The diagram in the next page shows the interdependence of the various results
proved in this paper. It does not refer to results of other authors; to such
results we refer in the proofs of our results.

Notation and conventions in the diagram of the next page

> T=Theorem P=Proposition L=Lemma
>
> ---
> ---
> means that “res-upp” is based on “res-low”.

---
2.1
---
4.1
---
4.2
---
4.4
---
4.3
---
3.3
---
4.5
---
3.3
---
4.7
---
4.6
---
3.3
---
5.1
---
3.3
---
7.1
---
7.2
---
5.3
---
3.4
---
3.8
---
5.3
---
3.14
---
3.11
---
3.12
---
3.13
---
5.2
---
7.3
---
7.4
---
3.4
---
3.9
---
7.5
---
7.6
---
5.5
---
3.17
---
3.15
---
3.16
---
3.10
---
5.4
---
7.7
---
7.8
---
7.9
---
7.10
---
5.7
---
3.20
---
3.18
---
3.19
---
5.6
---
7.11
---
7.12
---
7.13

## References

  * [1] W. Bosma, J. Cannon, C. Playoust, The Magma algebra system. I. The user language, J. Symbolic Comput. 24 (1997), 235–265.
  * [2] Y. Bugeaud, M. Mignotte, S. Siksek, _Classical and modular approaches to exponential Diophantine equations. I. Fibonacci and Lucas perfect powers_ , Ann. of Math. (2) 163 (2006), 969–1018.
  * [3] J. H. E. Cohn, _On square Fibonacci numbers_ , J. London Math. Soc. 39 (1964), 537–541.
  * [4] J. H. E. Cohn, _Square Fibonacci numbers, etc._ , Fibonacci Quart. 2 (1964), 109–113.
  * [5] J. Gebel, A. Pethő and H.G. Zimmer, _Computing integral points on elliptic curves_ , Acta Arith. 67 (1994), 171–192.
  * [6] M. Laurent, _Linear forms in two logarithms and interpolation determinants II_ , Acta Arith. 133 2008), 325–349.
  * [7] M. Laurent, M. Mignotte, Y. Nesterenko, _Formes linéaires en deux logarithmes et déterminants d’interpolation_ , J. Number Th. 55 (1955), 285-321.
  * [8] M. Mignotte, A. Pethő, _Sur les carrés dans certaines suites de Lucas_ , Journal de Théorie des Nombres de Bordeaux 5 (1993), 333–341.
  * [9] K. Nakamula, A. Pethő, _Squares in binary recurrence sequences_ , in “Number Theory”, Eds.: Győry/Pethő/Sós, Walter de Gruyter GmbH & Co., Berlin-New York 1998, pp. 409–421.
  * [10] The PARI Group, PARI/GP version 2.12.0, Univ. Bordeaux, 2019, http://pari.math.u-bordeaux.fr/.
  * [11] M. Pohst, H. Zassenhaus, _Algorithmic Algebraic Number Theory_ , Encyclopedia of Mathematics and its Applications, Cambridge University Press, Cambridge 1990.
  * [12] D. Poulakis _A simple method for solving the Diophantine equation $Y^{2}=X^{4}+aX^{3}+bx^{2}+cX+d$_, Elem. Math. 54 (1999), 32–36.
  * [13] P. Ribenboim, W.L. McDaniel _The square terms in Lucas sequences_ , J. Number Th. 58 (1996), 104–123.
  * [14] R.J. Stroeker and N. Tzanakis, _Solving elliptic Diophantine equations by estimating linear forms in elliptic logarithms_ , Acta Arith. 67 (1994), 177–196.
  * [15] N. Tzanakis, _Solving elliptic diophantine equations by estimating linear forms in elliptic logarithms. The quartic case_ , Acta Arithm. 75 (1996), 165–190.
  * [16] P. M. Voutier, _Primitive divisors of Lucas and Lehmer sequences, III_ , Math. Proc. Camb. Phil. Soc. 123 (1998), 407–419.
  * [17] O. Wyler, _Solution of problem 5080_ , Amer. Math. Monthly 71 (1964), 220–222.