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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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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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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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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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