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Find all four-digit positive integers $\overline{abcd}=10^3a+10^2b+10c+d$ $(a\ne0)$ such that: $\overline{abcd} =a^{a+b+c+d}-a^{-a+b-c+d}+a$

The only solution of the Diophantine equation $(1) \;\; \overline{abcd} = a^{a + b + c + d} - a^{-a + b - c + d} + a$, (where $\overline{abcd}$ is a four digit number) is $(a,b,c,d) = (2,0,1,8)$. Proof: Clearly the LHS of equation (1) is greater or equal to 1000, implying $a \neq 1$. Hence $a \geq 2$, which means the exponent $-a + b - c + d$ in equation (1) is non-negative, i.e. $(2) \;\; b + d \geq a + c$ Moreover by equation (1) $1000a \leq \overline{abcd} = a^{a+b+c+d} - a^{-a+b-c+d} + a < a^{a+b+c+d} + a$, yielding $(3) \;\; a^{a+b+c+d-1} > 999$. Futhermore $1000(a + 1) > \overline{abcd} = a^{a+b+c+d} - a^{-a+b-c+d} + a > a^{a+b+c+d} - a^{b+c+d} + a$ i.e. $(4) \;\; (a^a - 1) \cdot a^{b+c+d} \leq 999(a+1)$. By inequality (2) we obtain $a + b + c + d \geq 2(a + c) \geq 2a$, which combined with inequality (4) give us $(5) \;\; a^{2a} \leq 999(a+1)$. Hence $a \leq 3$ by inequality (5). Assume $a=3$. Then according to inequalities (3) and (4) we obtain $3^{b+c+d+2} > 999 > 3^6$ and ${\textstyle 3^{b+c+d} \leq \frac{3996}{26}} < 3^5$, yielding $b+c+d \geq 5$ and $b+c+d \leq 4$ respectively. This contradiction implies $a=2$. Applying the inequalities (3) and (4) we obtain $2^{b+c+d+1} > 999 > 2^9$ and $2^{b+c+d} \leq 999 < 2^{10}$, yielding $b+c+d \geq 9$ and $b+c+d \leq 9$ respectively. In other words, $b+c+d=9$. This means $-a+b-c+d = -(2 - c) + (9 - c) = 7 - 2c$, which inserted in equation (1) result in $\overline{2bcd} = 2000 + \overline{bcd} = 2^{2+9} - 2^{7-2c} + 2$, i.e. $(6) \;\; \overline{bcd} = 50 - 2^{7-2c}$. Thus we have $\overline{bcd} < 100$ by equation (6), which give us $b=0$. Consequently by equation (6) $(7) \;\; \overline{cd} = 50 - 2^{7-2c}$ with $(8) \;\; c+d=9$. According to equation (7) the digit $d$ is even. Thus $c$ is odd by equation (8). We also have $7 - 2c \geq 0$ by equation (7), i.e. $c \leq 3$. Therefore according to equation (7) $2^{7-2c} = 50 - \overline{cd} \geq 50 - 39 = 11 > 2^3$, yielding $7-2c>3$. Hence $c<2$, which means $c=1$ (since $c$ is odd) and $d=9-c=9-1=8$ by equation (8). Conclusion: The only solution of equation (1) is $(a,b,c,d) = (2,0,1,8)$. q.e.d.