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? Is there a solution to:
$a+b^m=b+c^n=c+a^l$ for l,m,n >1 and a, b, c distinct odd primes?
I've had a play around with specific possible solutions and there are lots of possibilities that may be systematically eliminated but I cannot see any obvious way to progress beyond specific cases. Is there any area of research that might be able to shed light on this, or is it a known result?
 A: Here is somewhat interesting feature of such primes.
Without loss of generality, there are two cases to consider:
1) If $a < b < c$ then $a^{\ell} < c^n < b^m$ and thus 
$$0 < c^n - a^{\ell} = c-b < c-a$$
2) If $b < a < c$ then $a^{\ell} < b^m < c^n$ and thus 
$$0 < c^n - b^m = a-b < c-b$$
That is, two primes out of $a,b,c$ must be such that the difference between their non-trivial powers is strictly smaller than their own difference. 
A: 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.
