Taking $\pmod{11}$ we get that $4^y \equiv 4,5,9,3,1 \pmod{11}$ and $x^5 \equiv \pm1, 0 \pmod{11}$, and since $11 \vert 2013$, we get that $4^y \equiv 1 \pmod{11} \implies 5\vert y$. So write $y = 5a$. Now note that $2013 = 3 \cdot 11 \cdot 61$. Consider a prime $p \vert x+4^a$. Clearly $p = 3, 11$ or $61$. But by LTE, $\upsilon_p(x^5+4^{5a}) = \upsilon_p(x+4^{a}) + \upsilon_p(5) = \upsilon_p(x+4^{a})$. So that means if any prime $p$ of $2013$ divides $x+4^a$, then all $z$ powers of $p$ in $2013^z$ divide $x+4^a$. $(*)$ Now note that the following inequality holds: \[(x+4^a)^5 > x^5+4^{5a} \ge \frac{(x+4^a)^5}{16}\] by Power-Mean inequality. So we have 2 cases: Case 1: The only prime dividing $x+4^a$ is $3$. Then $x+4^a \le 3^z$ so \[2013^z = x^5+4^{5a} < (x+4^a)^5 \le (3^z)^5 = 243^z,\] which is a contradiction. Case 2: A prime $p$ that is at least $11$ divides $x+4^a$. By the above $(*)$, $p^z \vert x+4^a \implies x+4^a \ge p^z \ge 11^z$. So \[2013^z = x^5+4^{5a} \ge \frac{(x+4^a)^5}{16} \ge \frac{(11^z)^5}{16} > 2013^z,\] which again is a contradiction. So there are no solutions.

Something else from Zsigmondy Theorem? All the solution was clear and well-explained but I don't know if Power-Mean inequality is a special case of a more general inequality, a kind of lemma...?

Particular case of Hölder $(1+1)(1+1)(1+1)(1+1)(x^5 + (4^a)^5) \geq (x + 4^a)^5$.

Thanks mavropnevma!!!

I have a bit different final result then mathocean97 mathocean97 wrote: Taking $\pmod{11}$ we get that $4^y \equiv 4,5,9,3,1 \pmod{11}$ and $x^5 \equiv \pm1, 0 \pmod{11}$, and since $11 \vert 2013$, we get that $4^y \equiv 1 \pmod{11} \implies 5\vert y$. So write $y = 5a$. Now note that $2013 = 3 \cdot 11 \cdot 61$. Consider a prime $p \vert x+4^a$. Clearly $p = 3, 11$ or $61$. But by LTE, $\upsilon_p(x^5+4^{5a}) = \upsilon_p(x+4^{a}) + \upsilon_p(5) = \upsilon_p(x+4^{a})$. So that means if any prime $p$ of $2013$ divides $x+4^a$, then all $z$ powers of $p$ in $2013^z$ divide $x+4^a$. $(*)$ in modulo 3 $x\equiv -1 (mod 3)$ so $x+4^k\equiv 0 (mod 3)$ and since $(x^5+(4^k)^5)/(x+4^k)$ is bigger then $(x+4^k)$ we have only 2 cases 1) $x+4^k=3^z$ then after plugging $x=3^z-4^k$ in original equation and canceling $3^z$ from both side we remain with conclusion that in modulo 3 $LHS\equiv 2 (mod 3)$ so $z$ is odd. But in modulo 8 we have $RHS\equiv 5 (mod 8)$ so $x^5\equiv x\equiv 5 (mod 8)$ but $x\equiv 3^z-4^k\equiv 3 (mod 8)$ we have contradiction. 2) $x+4^k=33^z$ then after plugging $x=33^z-4^k$ in original equation and canceling $33^z$ from both side we remain with conclusion that in modulo 3 $LHS\equiv 2 (mod 3)$ but $RHS\equiv 1 (mod3)$, contradiction so no solution.

Once we have proved that $v_p(x^5+4^{5a})=v_p(x+4^a)$ with p=3,11,61 we have that every prime that divides the RHS divdes $x+4^a$, so $x^4+...+4^{4a}$ can't be divided by any prime which divides the RHS so it must be 1, which yelds no solution. Is it wrong?

Yes, it's wrong. LTE only works in one direction; if $p\mid x+4^a$, then we can compute its power in $x^5 + 4^{5a}$. So all the following discussion is needed, in case the primes $3,11$ and $61$ are split between the two factors.

I used the mod11 and concluded in the fact that $y=5a$ then I used the LTE and knowing that $ \upsilon_3(x^5+4^{5a}) =\upsilon_3(x+4^{a})+\upsilon_3(5) =\upsilon_3(x+4^{a}) $. I was trying to fing the biggest power 3 that divides $x+4^a$ in order to restrict the possible values of $z$. Also why doesn't anybody check the possibility of 61 dividing $x^5+4^{5a}$. Last, why should $ x+4^a\le 3^z $. From the moment that $3^z$ divides $x+4^a$ ths means that $x+4^a>=3^z$.

Actually the solution of 'liimr' is the most beautiful and effect

Zsigmondy kills it rather quickly. Are contestants allowed to use it without a proof though?

I think there is no need of Zsigmondy in this problem, All we need LTE, however this also can be done by basic divisibility rules

I have another solution to the last part: $(a+b)(\frac {a^{5}+b^{5}} {a+b})=2013^{z}$ Now we will show the following inequalities $(a+b)<(a+b)^{2}\le \frac {a^{5}+b^{5}} {a+b}<(a+b)^{4}$ Since $a+b=x+4^{k}\ge 4$ we obviously have $(a+b)<(a+b)^{2}$ and from Cauchy-Shwarz inequality: $(a^{5}+b^{5})(a+b)\ge (a^{3}+b^{3})^{2}$ and by Power-Mean $a^{3}+b^{3}\ge \frac {(a+b)^{3}} {4}$ so we have $(a^{5}+b^{5})(a+b)\ge (a^{3}+b^{3})^{2}\ge (\frac {(a+b)^{6}} {16})\ge (a+b)^{4}$ (do not forget $a+b=x+4^{k}\ge 4$ ).And the last one $\frac {a^{5}+b^{5}} {a+b}<(a+b)^{4}$ is obvious Now from $gcd(a+b,\frac {a^{5}+b^{5}} {a+b})=1$,we have the following cases: 1)$a+b=3^{z}$ and $\frac {a^{5}+b^{5}} {a+b}=671^{z}$,but from above we must have $3^{4}>671$, contradiction 2)$a+b=11^{z}$ and $\frac {a^{5}+b^{5}} {a+b}=243^{z}$,but taking the original equation in $modulo 3$ we obtain $a=x\equiv 2(mod 3)$ and $b=4^{k}\equiv 1(mod 3)$ so $a+b\equiv 0(mod 3)$ and $11\not\equiv 0(mod3)$,contradiction 3)$a+b=33^{z}$ and $\frac {a^{5}+b^{5}} {a+b}=61^{z}$,but from above we must have $33^{2}<61$, contradiction Finally there are no solutions in positive integers

st_atan wrote: Zsigmondy kills it rather quickly. Are contestants allowed to use it without a proof though? What does the quick solution with Zsigmondy look like?

manuel153 wrote: st_atan wrote: Zsigmondy kills it rather quickly. Are contestants allowed to use it without a proof though? What does the quick solution with Zsigmondy look like? Why you should use Zsigmondy ? This statement: "But by LTE, $\upsilon_p(x^5+4^{5a}) = \upsilon_p(x+4^{a}) + \upsilon_p(5) = \upsilon_p(x+4^{a})$" can be proved using congruences almost straightforward as follows: As we seen in the start of the solution of mathocean97 we have $x \equiv -1 (mod 3), 4^y \equiv 1(mod 3), y=5a$, So we consequently have using $x^5 +4^{5a}=(x+4^a)(x^4-x^3.4^a+x^2.4^{2a}-x.4^{3a}+4^{4a})$ and $x^4-x^3.4^a+x^2.4^{2a}-x.4^{3a}+4^{4a} \equiv 1 -(-1)^3.1 + (-1)^2.1 - (-1).1+1 \equiv 5 (mod 3)$ that's why the above conclusion ...

Take the equation modulo 11, then since $11|2013$, \[x^5+4^y\equiv 0\pmod{11}.\] But $x^5\equiv \{0, 1, -1\}\pmod{11}$ since $n^{10}\equiv 1\pmod{11}$ by Euler's Theorem. This gives the solutions modulo $11$ to be $(x^5, 4^y)\equiv\{(0,0), (1, -1), (-1, 1)\}\pmod{11}$. However, $4^y\equiv\{4, 5, 9, 3, 1\}\pmod{11}$ and so the only possible solution is $4^y\equiv 1\pmod{11}$ which is true when $5|y$ since the order of 4 modulo 11 is 5. Therefore let $y=5b$. Now we have \[x^5+(4^b)^5=2013^z.\] But $x+4^b|x^5+(4^b)^5=2013^z$. Since we know that $x+4^b\neq 1$, we must have $x+4^b=3^i\cdot 11^j\cdot 61^k$ where $i,j,k\in \mathbb{Z}^{+}$. Now using LTE, we have \[v_3(2013^z)=z=v_3(x^5+(4^b)^5)=v_3(x+4^b),\] thus $3^z||x+4^b$. Likewise, $61^z||x+4^b$ and $11^z||x+4^b$. Clearly no other primes can divide $x+4^b$ so therefore $x+4^b=2013^z$. However, $2013^z=x^5+(4^b)^5>x+4^b=2013^z$ which is a contradiction. Therefore, there are no such solutions. $\blacksquare$

I think that the BMO jury must recognise the LIFTING THE EXPONENT LEMMA, so the contestants shouldn't write it's proof for p odd prime and for p=2, although the proof is very easy.

Does that mean the BMO jury currently does not? Why is that so?

I heard that some contestants who solved this by not involving the proof, didn't get all the maximal 10 points.

mavropnevma wrote: Yes, it's wrong. LTE only works in one direction; if $p\mid x+4^a$, then we can compute its power in $x^5 + 4^{5a}$. So all the following discussion is needed, in case the primes $3,11$ and $61$ are split between the two factors. Mavropnevma, can you explain a bit this? Because i agree with him. 3,11,61 doesnt divide $x^4+...+4^{4a}$ which is impossible since no other primes can divide this number. And note that JSGandora solved by this way. So everything look true. Please give response if im wrong or not.

JSGandora wrote: ... Therefore let $y=5b$. Now we have \[x^5+(4^b)^5=2013^z.\] But $x+4^b|x^5+(4^b)^5=2013^z$. Since we know that $x+4^b\neq 1$, we must have $x+4^b=3^i\cdot 11^j\cdot 61^k$ where $i,j,k\in \mathbb{Z}^{+}$. Now using LTE, we have ... What I've said pertains precisely to this mistake. Not all three exponents $i,j,k$ must be non-negative, as used in the above. Why if a prime $p \mid A^5 + B^5 = (A+B)(A^4 - A^3B + A^2B^2 - AB^3 + B^4)$, do we need have $p\mid A+B$ ? So, as $3,11,61 \mid 2013^z = x^5+(4^b)^5$, why do we need have $3,11,61 \mid x+4^b$ ? Can it not be that some of $3,11,61$ might divide $x^5+(4^b)^5$, but not $x+4^b$ ? A priori, it can be (and by Zsigmondy, it actually does, with the noted exceptions).

mavropnevma wrote: JSGandora wrote: ... Therefore let $y=5b$. Now we have \[x^5+(4^b)^5=2013^z.\] But $x+4^b|x^5+(4^b)^5=2013^z$. Since we know that $x+4^b\neq 1$, we must have $x+4^b=3^i\cdot 11^j\cdot 61^k$ where $i,j,k\in \mathbb{Z}^{+}$. Now using LTE, we have ... What I've said pertains precisely to this mistake. Not all three exponents $i,j,k$ must be non-negative, as used in the above. Why if a prime $p \mid A^5 + B^5 = (A+B)(A^4 - A^3B + A^2B^2 - AB^3 + B^4)$, do we need have $p\mid A+B$ ? So, as $3,11,61 \mid 2013^z = x^5+(4^b)^5$, why do we need have $3,11,61 \mid x+4^b$ ? Can it not be that some of $3,11,61$ might divide $x^5+(4^b)^5$, but not $x+4^b$ ? A priori, it can be (and by Zsigmondy, it actually does, with the noted exceptions). "So, as $3,11,61 \mid 2013^z = x^5+(4^b)^5$, why do we need have $3,11,61 \mid x+4^b$ ? Can it not be that some of $3,11,61$ might divide $x^5+(4^b)^5$, but not $x+4^b$ ?" you said. By Lte we have if $3^a|x^5+(4^b)^5$ , then $3^a|x+4^b$ because $v_p(2013^z)=v_p(x+4^b)=v_p(x^5+4^{5b})$ where p is 3 ,11 or 61. And if you read the prove you can see.I read it from resources page , Amir Hosseins article.So if im wrong again please explain a bit.LTE says exactly what i said.

Mathematicalx wrote: By Lte we have if $3^a|x^5+(4^b)^5$ , then $3^a|x+4^b$ ... No, no, and no again. You refuse to read what I'm saying, and quote LTE in reverse way. Do we have $41\mid 1^5 + 4^5 = 5^2\cdot 41$ ? Yes. Do we have $41\mid 1 + 4 = 5$ ? No. (This is not LTE) Do we have $5\mid 1 + 4 = 5$ ? Yes. Do we have $5^2\mid 1^5 + 4^5 = 5^2\cdot 41$ ? Yes. (This is LTE) If we don't make an effort, we'll never get to the end of this ...

Thanks, i understand now.

How do you get RHS = 5 (mod 8) ??

You probably refer to the following (when the post is so remote, please quote the relevant part) liimr wrote: ... we remain with conclusion that in modulo 3 $LHS\equiv 2 (mod 3)$ so $z$ is odd. But in modulo 8 we have $RHS\equiv 5 (mod 8)$ ... You missed the conclusion "$z$ is odd". Since $2013\equiv 5 \pmod{8}$ and $5^2\equiv 1 \pmod{8}$, it follows $2013^z \equiv 2013 \equiv 5 \pmod{8}$.

Another similar solution, We have $ (x+4^a)(x^4-x^3 \cdot 4^a+x^2 \cdot 4^{2a}-x \cdot 4^{3a}+4^{4a}) =2013^z $ and let $ d=gcd(x+4^a, (x^4-x^3 \cdot 4^a+x^2 \cdot 4^{2a}-x \cdot 4^{3a}+4^{4a}) $. Let us assume that $ d>1 $ so there exist prime $ p $ such that $ p|d $ and now $ 4^a\equiv -x\pmod{p} $ so $ x^4-x^3 \cdot 4^a+x^2 \cdot 4^{2a}-x \cdot 4^{3a}+4^{4a}\equiv 5x^4\equiv 0\pmod{p} $ but since $ p|RHS, p $ can not be $ 5 $, so $ p|x^4 $ and since $ p $ is prime, $ p|x $. Now $ p|x+4^a, p|4^a $, on the other hand $ p|2013^z $ and since $ gcd(4^a, 2013^z)=1 $ we must have $ p=1 $ which is not true so now $ d=1 $ and we can continue like all the other solutions.

What is LTE and Zsigmondy again, like what does it state?

We have $x^5 \equiv 0, \pm 1 \pmod{11}$, so $4^y \equiv 0 , \pm 1 \pmod{11}$, giving $y \equiv 0 \pmod{5}$. Set $y=5k$. We now have $x^5+(4^k)^5 = 2013^z$. If $p|x+4^k$, we know that $p=3,11,$ or $61$. By LTE, we have $v_p(2013^z) = z = v_p(x+4^k)+v_p(5)=v_p(x+4^k)$. This gives us that $x+4^k = 1,3^z,11^z$ or no less than $33^z$. Clearly the first is impossible. Meanwhile, from Power Mean Inequality, we have $$(x+4^k)^5 > x^5+(4^k)^5=2013^z > \frac{(x+4^k)^5}{16}$$Simple calculations show that $$16^z < x+4^k < \sqrt[5]{16} \cdot 25^z$$which is clearly a contradiction to the above argument.

Only solutions x=y=z=0 because we use mod 11

Solution: Zsigmondy overkills it! Observe that $x^5\equiv 0,1,-1\pmod {11}$. Also, $11|2013$. Consider $y=5k+r $, $0\le r<5$. Case checking shows that $r=0\implies 5|y $. Let $y=5k $. So, we have, $x^5+(4^k)^5=2013^z $. Applying Zsigmondy's Theorem, we find that there exists a primitive prime factor, $p $, of $x^5+(4^k)^5$. This gives $p|2013\implies p=3,11,61$. But, $2013^z=x^5+(4^k)^5=(x+4^k)(x^4-x^34^k+.....+256^k) $. This implies that $x+4^k $ has atleast one of $3,11,61$ as its prime factor. This contradicts $p$ being primitive. Also, the exception doesn't hold in this case. Hence, there are no natural solutions to the equation.

We see x must be odd but: \(RHS \equiv 1 \pmod 4\) so \(x^5\equiv 1 \pmod 4 \Rightarrow x\equiv 2 \pmod 4 \Rightarrow 2|x \Rightarrow CONTRADICTION\) Isn't this simpler? (assuming it's correct)

How can we apply zsigmondy theorem here we don't know whether $x>4^{k}$

Superguy we don't need x>4^k. It works just fine if x<4^k

Apologies for bumping this, but I did not see a full simple solution avoiding LTE/Zsigmondy, so let's post one. Motivated by $a^{\frac{p-1}{2}} \equiv 0,\pm 1 \pmod p$ for a prime $p$ (follows by Fermat's little theorem), let's begin with mod $11$ - we must necessarily have $4^y \equiv 0, \pm 1 \pmod {11}$ and since $4^1 \equiv 4$, $4^2 \equiv 5$, $4^3 \equiv 9$, $4^4 \equiv 3$ and $4^5 \equiv 1$, we must have $y=5a$ for some positive integer $a$. Now let $B=4^a$. Then $2013^z = x^5 + 4^{5a} = (x+B)(x^4 - x^3B + x^2B^2 - xB^3 + B^4)$. Observe that the factors on the right are coprime - as if $p$ is a common prime divisor of both, then $B\equiv -x \pmod p$ and the second factor gives $5x^4 \equiv 0 \pmod p$; however $p$ divides $2013$, so $p$ cannot be $2$ or $5$ and so $p\mid x$, whence $p\mid B = 4^a$, contradiction! Having in mind $2\leq x+B < x^4 - x^3B + x^2B^2 - xB^3 + B^4$ for except for $x=B=1$ (to prove that, check quickly $x=B$ and now without loss of generality let $x>B$ - then the inequality is equivalent to $x^3(x-B) + x(B^2(x-B)-1) + B^4 - B > 0$ which is true), we deduce that $x+B = 3^z$ or $x+B = 11^z$ or $x+B = 33^z$. If $x+B \geq 11^z$, then by the Power-Mean inequality we have $2013^z = x^5 + B^5 \geq \frac{(x+B)^5}{16} \geq \frac{11^{5z}}{16}$, contradiction; while if $x+B = 3^z$, then $2013^z = x^5 + B^5 < (x+B)^5 = 243^z$, contradiction. Therefore there are no solutions.

First we start with a claim - Key observation is that $2013= 3.11.61$ Claim - $5 \mid y$ Proof - For this note that $x^{5} \equiv -4^{y} \mod 11$ and $$-4^{y} \equiv 7 , 6 , 2 , 8, 10 \mod 11$$and as $$x^{5} \equiv 1 , -1 \mod 11$$, so only thing common in both residues is $-1$ , giving $5 \mid y$ . Now let $y = 5k$ , so we can rewrite the equation as $x^5 + 4^{5k} = 2013^z$ , so any prime dividing $x + 4^k$ must be from the set ${3, 11, 61}$ , also if $p \mid x + 4^k$ , then $\nu_p({x + 4^k}) = z$ (say from LTE) as $gcd(5 , 3.11.61) = 1$ , so if $2$ primes divide $x + 4^k$ , then we get $$ x + 4^k \ge 3^{z}. 11^{z}$$, but from Power mean inequality we have $$ \frac{x^5 + 4^{5k}}{2} \ge \frac{(x^5 + 4^k)^5}{32} \longrightarrow 2013^z \ge \frac{33^{5z}}{16}$$which obviously causes size issues. So $x + 4^k = p^z$ where $p \in {3, 11, 61}$ , but note that again from Power Mean Inequality , we get that $$ 3^z . 61^z . 11^z \ge \frac{p^{5z}}{16}$$, which gives $p = 3$ , but this fails as this gives(from Binomial Theorem) $$(x+4^k)^5 > x^5 + 4^{5k} = 3^z . 61^z . 11^z > 3^{6z} = (x+4^{k})^6$$, a contradiction. Hence we have no solutions. $\blacksquare$

My solution contains: $mod11$,$x+4^k=3^a11^b61^c$,lte contradicition like others.