# 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?

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If you keep a, b and c fixed, log a/log b is well approximated by m/l. Making that a linear form in logarithms, that should force l and m to be fairly big. Which is less plausible than solutions at random for small values. This is an exponential Diophantine equation, but not hopeless. – Charles Matthews Jun 1 '10 at 14:24
I added tag "diophantine-equations". A nice sourse for exponential DEs is the book of Shorey and Tijdeman. I've not seen your equation, but as Charles says it is hardly hopeless. But most probably routine. – Wadim Zudilin Jun 1 '10 at 14:51

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.

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Thanks. That is an interesting way of demonstrating how rare such numbers must be. – Francis Davey Jul 17 '10 at 6:41

for above question,Is there a solution to: $a+b^m=b+c^n=c+a^l$ for l,m,n >1 and a, b, c distinct positive integers?

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Yes, that would be a more unconstrained question, which I also don't know the answer to. – Francis Davey Jun 23 '10 at 7:19

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.

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Thanks. I'll have to go away and think about that. I'm not sure what you mean by "first equation: similar to z[i]". – Francis Davey Aug 20 '10 at 7:36
First question - are you assuming that m, n are even? If they are not then $b^m/2$ won't be an integer. Second, where does the assumption that l, m and n are greater than 1 enter? – Francis Davey Aug 20 '10 at 11:02
m,n,l are greater than 1,and $b_1$ and $a_1$ may be irrational or rational numbers,$z[i]$ is about integers,properties of this method is similar to $z[i]$,not the same – Hashem sazegar Aug 20 '10 at 11:27