Math-lover123 wrote: Solve the following diophantine equation in the set of nonnegative integers: $11^{a}5^{b}-3^{c}2^{d}=1$. Case I. If $c=0$ then $2^d+1=11^a5^b$. If $b=0$ then $11^a-1=2^d$. In the case $a$ is odd then by LTE we obtain $v_2(11^a-1)=v_2(11-1)=1=d$, then $11^a=3$, a contradiction. Therefore $a$ is even. We also have by LTE \[v_2(11^a-1)=v_2(11-1)+v_2(11+1)+v_2(a)-1=2+v_2(a)=d\] It is easy to prove by induction that $2^m \ge m+1$ implies $a \ge v_2(a)+1$ or $a+1 \ge d$. Hence $11^a-1=2^d \le 2 \cdot 2^a$ or $2 \cdot 2^a+1 \ge 11^a$. It follows that $a=0$ so $2^d=0$, a contradiction. So there is no solution in this case. If $a=0$ then $5^b-1=2^d$. We have $b \ge 1$. If $b$ is odd then by LTE we have $v_2(5^b-1)=v_2(5-1)=2=d$. It follows that $5^b=5$ implies $b=1$. Thus, $(a,b,c,d)=(0,1,0,2)$. If $b$ is even then by LTE we have \[v_2(5^b-1)=v_2(5^2-1)+v_2(b)-1=2+v_2(b)=d.\] Similarly, we get $b \ge v_2(b)+1$ so $b+1 \ge d$ implies $5^b-1 \le 2 \cdot 2^b$. We obtain $b=0$, a contradiction since $b \ge 1$. So, in this case we get $(a,b,c,d)=(0,1,0,2)$. If $a,b \ge 1$ then $5|2^d+1$ and $11|2^d+1$. From $5|2^d+1$ we get $2 \parallel d$. But from $11|2^d+1$ we get $2 \nmid d$, a contradiction. Case II. If $d=0$ then $3^c+1=11^a5^b$, a contradiction since $3^c+1 \equiv 0 \pmod{2}$ but $11^a5^b \equiv 1 \pmod{2}$. Case III. If $d=1$ then $2 \cdot 3^c+1=11^a5^b$. It is easy to see that $c \ge 1$. If $b=0$ then $2 \cdot 3^c+1=11^a$. Since $v_2(11^a-1)=1$ so $a$ is odd, a contradiction because we have $11^a-1 \equiv 1 \pmod{3}$ then. Do $b \ge 1$ Similarly, we get $a \ge 1$. Hence $5|2 \cdot 3^c+1$ and $11| 2 \cdot 3^c+1$. We have two tables: \[\begin{array}{|c|c|} \hline c \pmod{4} & 2 \cdot 3^c+1 \pmod{5} \\\hline 0 & 3 \\\hline 1 & 2 \\\hline 2 & 4 \\\hline 3 & 0 \\\hline \end{array}\] Thus, $c \equiv 3 \pmod{4}$. Next: \[\begin{array}{|c|c|} \hline c \pmod{5} & 2 \cdot 3^c+1 \pmod{11} \\\hline 0 & 3 \\\hline 1 & 7 \\\hline 2 & 8 \\\hline 3 & 0 \\\hline \end{array}\] Thus, $c \equiv 3 \pmod{5}$. We obtain $c \equiv 3 \pmod{20}$. Let $c=20c_1+3 \; (c_1 \in \mathbb{N})$. The equation is equivalent to \[55 \left( 11^{a-1}5^{b-1}-1 \right)=54 \left( 3^{20c_1}-1 \right).\] If $b \ge 2$ then $5 \parallel 55 \left( 11^{a-1}5^{b-1}-1 \right)$ but $5^2|3^{20c_1}-1$, a contradiction. Thus, $b=1$. Therefore $55 \left( 11^{a-1}-1 \right)=54 \left( 3^{20c_1}-1 \right)$. But $11^2|3^{20c_1}-1$ and so, we get $a=1$. It follows that $c=3$. Thus, $(a,b,c,d)=(1,1,3,1)$. Case IV. If $d \ge 2,c \ge 1$. We have $4|3^c2^d$ and $5^b \equiv 1 \pmod{4}$ so $11^a \equiv 1 \pmod{4}$. It follows that $a \equiv 0 \pmod{2}$ implies $11^a \equiv 1 \pmod{3}$. We also have $3^c2^d +1 \equiv 1 \pmod{3}$ so $5^b \equiv 1 \pmod{3}$. Thus, $b \equiv 0 \pmod{2}$. Let $a=2a_1,b=2b_1 \; (a_1,b_1 \in \mathbb{N})$. We have \[\left( 11^{a_1}5^{b_1}-1 \right) \left( 11^{a_1}5^{b_1}+1 \right)=3^c2^d.\] Since $\left( 11^{a_1}5^{b_1}+1 \right)- \left( 11^{a_1}5^{b_1}-1 \right)=2$ so we have some small cases: Case 1. If $\begin{cases} 11^{a_1}5^{b_1}-1=2 \cdot 3^c \\ 11^{a_1}5^{b_1}+1=2^{d-1} \end{cases}$ then $2^{d-1}-2 \cdot 3^c=2$ or $2^{d-2}-3^c=1$. We find $d=2,c=1$ so $11^a5^b=3 \times 2^2+1=13$, a contradiction. Case 2. If $\begin{cases} 11^{a_1}5^{b_1}-1=2^{d-1} \\ 11^{a_1}5^{b_1}+1=2 \cdot 3^c \end{cases}$ then $2 \cdot 3^c-2^{d-1}=2$ or $3^c-2^{d-2}=1$ implies $c=1,d=3$. So $5^a11^b=3^1 \times 2^3+1=5^2$ implies $a=0,b=2$. Thus, $(a,b,c,d)=(0,2,1,3)$. Thus, $\boxed{(a,b,c,d)=(1,1,3,1),(0,1,0,2),(0,2,1,3)}$.

The （0，2，1，3）is a solution too。

lin wrote: The （0，2，1，3）is a solution too。 Thank you, I edited my solution.