I need help to solve the following Diophantine equation:$$ 12^x + y^4 = 56^z $$
where $x,y,z$ are non-negative integers.
My attempt: We can see immediately that $x=y=z=0$ is a solution. By taking mod $3$ we have: $$y^4 \equiv 2^z \text{ (mod 3)}$$ But we already know that $y^4 \equiv 0 \text{ or } 1 \text{ (mod 3)}$ and $2^z$ is not divisible by $3$, therefore $2^z \equiv 1 \text{ (mod 3)}$, so $z$ is even. Let $z=2t$ , we then subtract $y^4$ from both sides to get $$12^x = \left( 56^t-y^2 \right) \left( 56^t+y^2 \right)$$
So we have $56^t-y^2= 2^a3^b$ and $56^t+y^2 = 2^{2x-a}3^{x-b}$ for some non-negative intgers $a,b$. But since $\left( 56^t-y^2 \right) + \left( 56^t+y^2 \right)$ is not divisible by $3$, only one of the two terms is divisible by $3$, which means $b = 0 \text{ or } b=x$.
This is where I got stuck, I don't know what to do next. Am I going the wrong direction? Please help me.
Best Answer
The only solution is $x=y=z=0$.
If $x=0$, then $1+y^4=2^{3z}7^z$. If $z=0$, then $y=0$. If $z\geqslant 1$, then $y^4\equiv 6\pmod 7$ which is impossible.
In the following, $x\geqslant 1$.
You've already proved that if $x\geqslant 1$, then $b=0$ or $b=x$.
If $b=0$, then it follows from $56^t-y^2= 2^a$ and $56^t+y^2 = 2^{2x-a}3^{x}$ that $$(2\cdot 56^t=)\ 2^{3t+1}7^t=2^a+2^{2x-a}3^x\tag1$$ (Case 1) If $a=x$, then dividing the both sides of $(1)$ by $2^x$ gives $2^{3t-x+1}7^t=1+3^x$. Supposing that $3t-x+1\geqslant 3$ gives $3^x\equiv 7\pmod 8$ which is impossible. If $3t-x+1=0$, then $-1=3\cdot 27^t-7^t$ which is impossible since RHS is positive. If $3t-x+1=1$, then $2=7^{-t}+(27/7)^t$ which is impossible since if $t=0$, then $x=0$, and if $t\geqslant 1$, then RHS is larger than $2$. If $3t-x+1=2$, then $12=3\cdot 7^{-t}+(27/7)^{t}$ which is impossible since if $t\geqslant 2$, then RHS is larger than $12$ with $t\not=0$ and $t\not=1$.
(Case 2) If $a\lt x$, then dividing the both sides of $(1)$ by $2^a$ gives $2^{3t-a+1}7^t=1+2^{2(x-a)}3^x$. Since RHS is odd, one has to have $3t-a+1=0$ implying $7^{t}=1+2^{2x-6t-2}3^x$. Since $(2(x-a)=)\ 2x-6t-2\geqslant 2$, one has $(-1)^t\equiv 1\pmod 4$, so $t$ is even. Setting $t=2u$, one has $(7^u-1)(7^u+1)=2^{2x-6t-2}3^x$. So, there are integers $m(\geqslant 1),n$ such that $7^u-1=2^{2x-6t-2-m}3^{x-n}$ and $7^u+1=2^m3^n$. Supposing that $n\geqslant 1$ gives $1^u+1\equiv 0\pmod 3$ which is impossible. So, $n=0$ and $(7^u+1-(7^u-1)=)\ 2=2^m-2^{2x-6t-2-m}3^{x}$. Supposing that $m\geqslant 2$ and $2x-6t-2-m\geqslant 2$ gives $2\equiv 0\pmod 4$ which is impossible. So, one has to have either $m=1$ or $2x-6t-2-m=1$. If $m=1$, then $2^{2x-6t-3}3^{x}=0$ which is impossible. If $2x-6t-2-m=1$, then $2^{m-1}-3^{x}=1$. Supposing that $m-1\geqslant 3$ gives $3^x\equiv 7\pmod 8$ which is impossible. So, $m-1\leqslant 2$, and so $(m-1,x)=(1,0),(2,1)$ for which $t$ cannot be an integer.
(Case 3) If $a\gt x$, then dividing the both sides of $(1)$ by $2^{2x-a}$ gives $2^{3t+1-2x+a}7^t=2^{2a-2x}+3^x$. Since RHS is odd, it follows that $3t+1-2x+a=0$ to have $7^t=4^{x-3t-1}+3^x$. Multiplying the both sides by $4^{3t+1}$ gives $4^{3t+1}\cdot 7^t=4^x+3^x\cdot 4\cdot 64^t$ from which one has $0\equiv 4^x(1+4(-1)^x)\pmod 7$ which is impossible.
If $b=x$, then it follows from $56^t-y^2= 2^a3^b$ and $56^t+y^2 = 2^{2x-a}3^{x-b}$ that $(2\cdot 56^t=)\ 2^{3t+1}7^t=2^a3^x+2^{2x-a}$ which has no solution because $(1)$ has no solution.
In conclusion, the only solution is $x=y=z=0$.