If $p$ is odd then it's easy to see max power of prime of $a,b$ are equal. So finally we've a equation of form $a^p+b^p=p^c$ and $GCD(a,b)=1$ We know if $p$ divides $a+b,\frac {a^p+b^p}{a+b}$ then $p$ divides both of $a,b$ can't possible.(if $a,b>1$) So $p=2$ We've $a^2+b^2=2^c$ WLOG we can take both of $a,b$ odd. So we must have $a^2+b^2=2$ so $a=b=c=1,p=2$ If $b=1$ then we've $a^p+1=p^c$ Now it's easy to see $a+1|p$ so $a+1=p$ also $a^p+1=p(p^2.k+p)$ so $c\leq 2$ If $c=2$ then $p=3,a=2$

WakeUp wrote: Let $p$ be a prime number. Find all positive integers $a,b,c\ge 1$ such that: \[a^p+b^p=p^c.\] Strange, as I remember I solved this problem earlier. It is just a generalisation of BxMO 2010/4. This one is easy to solve by LTE. So if $p$ is odd, we just take $gcd(a,b)=1$ and by Fermat: $a^p \equiv a \pmod{p}$ so $p|a+b$ and $v_p(a^p+b^p)=v_p(a+b)+1$ Hence $a^p+b^p=p(a+b)$ and for $p\ge 5$ we see $a^p+b^p \ge 2(\frac{a+b}{2})^p>p(a+b)$ as $(\frac{a+b}{2})^{p-1}>2^{p-1}>p.$ Then just $p=3$ (BxMO-question) Remark you already forgot $(2,1,3,2).$ or $p=2$ and this is solved in post above.

Some of the solutions here are nice http://www.artofproblemsolving.com/Forum/viewtopic.php?f=56&t=468898

it's peace of a cake by LTE we can understand that a+b=p^(c-1) and then we use inequality !!!!! too easy as a TST

subham1729 wrote: We know if $p$ divides $a+b,\frac {a^p+b^p}{a+b}$ then $p$ divides both of $a,b$ can't possible.(if $a,b>1$) Totally absurd. If it were true, LTE would be false. Numerical example: $5|2+3;\; 5|\dfrac {2^5+3^5}{5}=55$

I think Zigmondies theorem helps here. We have $p|a+b (p\neq 2)$ But by the theorem there exists a primitive prime divisor of $a^p+b^p$ .Hence we should have $a=2,b=1$ and $p=3$

yeah MMEEvN this is too easy by Zsigmondys theorem: as $a+b$ is divisible by $p$ it has prime divisor other then $p$. So for special cases in Zsigmondys theorem $a=2$, $b=1$ and $p=3$, $c=2$

WakeUp wrote: Let $p$ be a prime number. Find all positive integers $a,b,c\ge 1$ such that: \[a^p+b^p=p^c.\] Immediate by Zsigmondy. $a^n+b^n$ has at least two prime divisors when $a>b>0, n\ge 2, (a,b,n)\neq (2,1,3)$. So here we have $a=b\,\Rightarrow\, (a,b,p,c)=(1,1,2,1)$ or $(a,b,p)=(2,1,3)\,\Rightarrow\, c=2$. liimr wrote: yeah MMEEvN this is too easy by Zsigmondys theorem: as $a+b$ is divisible by $p$ it has prime divisor other then $p$. So for special cases in Zsigmondys theorem $a=2$, $b=1$ and $p=3$, $c=2$ Also $a=b$ is too a special case.

Nice problem

related: https://artofproblemsolving.com/community/c6h1590647 (Bulgaria 1994, round ?) https://artofproblemsolving.com/community/c6h468898 (French 2012 TST q3) https://artofproblemsolving.com/community/c6h541628 (india 1995) https://artofproblemsolving.com/community/c6h144407 https://artofproblemsolving.com/community/c6h67350 (india 1995) https://artofproblemsolving.com/community/c6h478498 https://artofproblemsolving.com/community/c6h414864 https://artofproblemsolving.com/community/c6h128127 (india 1995) https://artofproblemsolving.com/community/c6h105949 https://artofproblemsolving.com/community/c6h87904 (Bulgaria 1994) https://artofproblemsolving.com/community/c6h111374 https://artofproblemsolving.com/community/c6h56830 https://artofproblemsolving.com/community/c6h32361 some sources say bulgaria , 1994 and india, 1995 but i cant verify both.

randomasdf97 has a very good solution. As for the case a=b and p=2, I think there are infinitely many solutions. We take a^2=2^c-1, and clearly we got infinitely many solutions.

subham1729 wrote: If $p$ is odd then it's easy to see max power of prime of $a,b$ are equal. So finally we've a equation of form $a^p+b^p=p^c$ and $GCD(a,b)=1$ We know if $p$ divides $a+b,\frac {a^p+b^p}{a+b}$ then $p$ divides both of $a,b$ can't possible.(if $a,b>1$) So $p=2$ We've $a^2+b^2=2^c$ WLOG we can take both of $a,b$ odd. So we must have $a^2+b^2=2$ so $a=b=c=1,p=2$ If $b=1$ then we've $a^p+1=p^c$ Now it's easy to see $a+1|p$ so $a+1=p$ also $a^p+1=p(p^2.k+p)$ so $c\leq 2$ If $c=2$ then $p=3,a=2$ "WLOG we can take both of $a,b$ odd." Why can you take this WLOG? Actually you can't, the problem where $a,b$ are even is another solution to the problem.

Case 1: $a=b$ Since thr LHS is even we have that $p=2$ and the diophane becomes $a^2=2^{c-1}$ thus $a=2^{\frac{c-1}{2}}$ and we see that $c$ is odd. Now the solution on this case is $(a,b,c,p)=(2^{\frac{c-1}{2}}, 2^{\frac{c-1}{2}},c,2)$ Case 2: $a \ne b$ W.L.O.G $a>b$ then assume that $(a,b,p) \ne (2,1,3)$ thus by Zsimondy we have that there exists a prime $q$ such that $q \mid a^p+b^p$ and $q \not \; \mid a+b$ and with this we have that $q \mid p^c$ meaning that $q=p$. Assume that $kp^{\alpha}=a$ and $lp^{\beta}=b$ where $\gcd(k,p)=\gcd(l,p)=1$ then replacing we have $p \beta<c$ and $k^pp^{\alpha-\beta}+l^p=p^{c-p \beta}$ thus $p \mid l$ and we have a contradiction!!. Thus we have $p \not \; \mid a,b$ thus by FLT: $a+b \equiv 0 \pmod p$ and we get contradiction!!. Thus we have $(a,b,p)=(2,1,3)$ meaning that $c=2$. Thus all the solutions are: $(a,b,c,p)=(2^{\frac{c-1}{2}}, 2^{\frac{c-1}{2}},c,2), (2,1,2,3), (1,2,2,3)$ And we are done

My answers are (a,b,c,p) (3^a,2*3^a,3*a+2,3) (2*3^a,3^a,3*a+2) (2^a,2^a,2*a+1,2) Its infinite many solutions