nt.number theory - Is there a solution to the a+b^m=b+c^n=c+a^l for l,m,n >1 and a, b, c distinct odd primes?

assume that $a>b>c$,and $a,b,c$ are distinct odd primes,we write above equation to form:

$c^n=(b^{m/2}+i(a-b)^{1/2})(b^{m/2}-i(a-b)^{1/2})$ and

$a^l=(c^{n/2}+i(b-c)^{1/2})(c^{n/2}-i(b-c)^{1/2})$

first equation: similar to $z[i]$,if $gcd(b^{m/2}+i(a-b)^{1/2},b^{m/2}-i(a-b)^{1/2})=d$,since

$NORM(2b^{m/2})=4b^m $and $NORM(b^{m/2}+/-(a-b)^{1/2})=b^m+(a-b)=c^n$ ,then $d=1$

so $b^{m/2}+i(a-b)^{1/2}=(a_1+ib_1)^n$ and $b^{m/2}-i(a-b)^{1/2}=(a_1-ib_1)^n$ ,so

$c=a_1^2+b_1^2$ and also $b^{m/2}+i(a-b)^{1/2}=r^n(cos(nx)+isin(nx))$ that

$r^2=a_1^2+b_1^2$,and$tan(x)=b_1/a_1$,then $b^{m/2}=c^{n/2}cos(nx)$ and $(a-b)^{1/2}=c^{n/2}sin(nx)$

with similar method from second equation ,we have:

$c^{n/2}=a^{l/2}cos(ly)$ and $(b-c)^{1/2}=a^{l/2}sin(ly)$

if we assume that $cos(nx)=cos(ly)$ so $c^{2n}=b^ma^l$ ,this is contradiction so above

equation has not any solution.