How to solve this equation $a^2+3b^2c^2=7^c$ Let $a,b,c$ be poistive integers,and such $\gcd(a,b)=\gcd(b,c)=\gcd(a,c)=1$,fine the all $a,b,c$ such
$$a^2+3b^2c^2=7^c$$
I'm not sure that this question has been studied, but I've been trying for a long time$(a,b,c\le 100)$, and there's only one set of solutions：$(a,b,c)=(2,1,1)$,But I can't prove it. I may need your help. Thank you
 A: We work in $\mathbb{Z}[\omega]$ where $\omega=\frac{1+i\sqrt{3}}2$. It is a factorial ring, and we factorize both sides as $(a+i\sqrt{3}bc)(a-i\sqrt{3}bc)=(2+i\sqrt{3})^n(2-i\sqrt{3})^n$. Since $2+i\sqrt{3},2-i\sqrt{3}$ are prime (and coprime), the guy $a+i\sqrt{3}bc$ can not be divisible by both (otherwise it is divisible by $(2+i\sqrt{3})(2-i\sqrt{3})=7$ that contradicts to our assumption that $\gcd(a,b)=\gcd(a,c)=1$), we see that $a+i\sqrt{3}bc=\varepsilon(2\pm i\sqrt{3})^c$ where $\varepsilon$ is a unit in $\mathbb{Z}[\omega]$. If $\varepsilon\ne \pm 1$, we get $\varepsilon=\frac{\pm 1\pm i\sqrt{3}}2$ and therefore $$\varepsilon(2\pm i\sqrt{3})^c= \left(\varepsilon (\pm i\sqrt{3})^n+(\text{something in } \mathbb{Z}[i\sqrt{3}])\right)\notin  \mathbb{Z}[i\sqrt{3}],$$
a contradiction.
So we get $\varepsilon=\pm 1$ and $a+i\sqrt{3}bc=\pm (2\pm i\sqrt{3})^c$. Consider two cases.

*

*$c$ is even, $c=2^\alpha \beta$ for odd $\beta$. Then
$$2i\sqrt{3}bc=\pm\left((2+i\sqrt{3})^{c}-(2-i\sqrt{3})^{c}\right)$$
is divisible by
$$
(2+i\sqrt{3})^{2^\alpha}-(2-i\sqrt{3})^{2^\alpha}=2i\sqrt{3}\cdot 4\\ \cdot((2+i\sqrt{3})^{2}+(2-i\sqrt{3})^{2})
((2+i\sqrt{3})^{4}+(2-i\sqrt{3})^{4})\ldots 
((2+i\sqrt{3})^{2^{\alpha-1}}+(2-i\sqrt{3})^{2^{\alpha-1}})
$$
which is divisible by $2i\sqrt{3}\cdot 2^{\alpha+1}$, thsu $b$ is also even, a contradiction with $\gcd(b,c)=1$.


*3 divides $c$. Since $(2+i\sqrt{3})^3-(2-i\sqrt{3})^3=18i\sqrt{3}$, we analogously get that 3 divides $b$ (the guys $A^2+AB+B^2$ for $A,B=(2\pm i\sqrt{3})^{3^t}$ are divisible by 3.)


*$c$ is odd and not divisible by 3. Let $3<p_1<p_2<\ldots<p_k$ be distinct prime divisors of $c$ (over $\mathbb{Z}$). Then $p_1$ divides (in $\mathbb{Z}[\omega]$) the number $(2+i\sqrt{3})^c-(2-i\sqrt{3})^c=\pm 2i\sqrt{3}bc$. In particular $p_1\ne 7=(2+i\sqrt{3})(2-i\sqrt{3})$.
3.1) If $p_1$ is prime in $\mathbb{Z}[\omega]$, then by Lagrange theorem it divides $(2 \pm i\sqrt{3})^{p_1^2-1}-1$, thus it divides $(2 + i\sqrt{3})^{p_1^2-1}-(2 - i\sqrt{3})^{p_1^2-1}$. Therefore $p_1$ divides $(2 + i\sqrt{3})^{\gcd(p_1^2-1,c)}-(2 - i\sqrt{3})^{\gcd(p_1^2-1,c)}=2i\sqrt{3}$, a contradiction. (We used that $p_1^2-1=4\cdot \frac{p_1-1}2\cdot \frac{p_1+1}2$ has prime divisors less than $p_1$, thus is coprime with $c$.)
3.2) If $p_1=q\bar{q}$ for a prime $q\in \mathbb{Z}[\omega]$, then by Lagrange theorem $q$ divides $(2 + i\sqrt{3})^{p_1-1}-(2 - i\sqrt{3})^{p_1-1}$ and we analogously get a contradiction.
