Solving Diophantine Equations Involving Exponentials as Positive Integers

Question

Find all positive integers $x$, $y$ and $z$ such that $$x^5+4^y=2013^z$$ Now as $$11|2013 \implies 11|4^y+x^5$$ Now by Euclid's Division Lemma we have $y=5q, 5q+1, 5q+2, 5q+3, 5q+4$, so we get $$4^y \equiv 1, 4, 5, -2 or 3 \mod 11$$ and also by Fermat's Little Theorem we have $x^{10} \equiv 1 \mod 11 \implies x^5 \equiv 1, -1 \mod11$, so the only possibility of $11|4^y+x^5$ is when $4^y \equiv 1 \mod11$ and $x^5 \equiv -1 \mod 11$ This gives $y=5q$ so our original equation becomes $$x^5+4^{5q}=2013^z \implies (x+4^q)(x^4-x^34^q+x^24^{2q}-4^{3q}x+4^{4q})=2013^z$$, Now as we know that $$(\frac{a^p+b^p}{a+b}, a+b)=1 or p$$ for an odd prime $p$ so we get $$(x+4^q, x^4-x^34^q+x^24^{2q}-4^{3q}x+4^{4q})=1 or 5$$ but as $5$ does not divide $2013^z$ so there gcd is $1$ So there are six possibilities $(l) x+4^q=11^z, x^4-x^34^q+x^24^{2q}-4^{3q}x+4^{4q}=183^z$ $(ll)x+4^q=183^z, x^4-x^34^q+x^24^{2q}-4^{3q}x+4^{4q}=11^z$ $(lll) x+4^q=33^z, x^4-x^34^q+x^24^{2q}-4^{3q}x+4^{4q}=61^z$ $(lV) x+4^q=61^z, x^4-x^34^q+x^24^{2q}-4^{3q}x+4^{4q}=33^z$ $(V) x+4^q=3^z, x^4-x^34^q+x^24^{2q}-4^{3q}x+4^{4q}= 671^z$ $(Vl) x+4^q= 671^z,x^4-x^34^q+x^24^{2q}-4^{3q}x+4^{4q}=3^z$ Now if $x>4^q \implies x \geq 4^q+1 \implies x-4^q \geq 1$ then we get $$ x^4-x^34^q+x^24^{2q}-4^{3q}x+4^{4q}=x^3(x-4^q)+4^{2q}x(x-4^q)+4^{4q} \geq x^3+4^{2q}x+4^{4q} > x^2+2x4^q+4^2q=(x+4^q)^2$$ and similar is the case of $4^q>x$ so the cases$(ll), (lll), (lV), (Vl)$ are not possible Now in case $(l)$ we have $$x+4^q=11^z, x^4-x^34^q+x^24^{2q}-4^{3q}x+4^{4q} =183^z$$ so we get $$(x+4^q)^4- (x^4-x^34^q+x^24^{2q}-4^{3q}x+4^{4q})=5x^34^q+5x^24^{2q}+54^{3q}x=11^{4z}-183^z \implies 5|11^{4z}-183^z$$ Now as the last digit of $11^{4z}$ is $1$ and the last digit of $183^z$ can be $3, 9, 7, 1$ and as $5|11^{4z}-183^z$ so the last digit of $183^z$ is $1$ so we get $4|z$ and so $z$ is even so we get $11^z \equiv 1 \mod 3$ also we have $4^q \equiv 1 \mod 3$ so we get $11^z-4^q \equiv 0 \mod 3$ and so $3|11^z-4^q$ but $11^z-4^q=x$ so $3|x$ but from our original equation we have $$x^5+4^y=2013^z$$ Now as $3|2013^z$ also $3|x$ so we get $3|4^y$ a contradiction So we are left with the case $(V)$ which gives $$x+4^q=3^z, x^4-x^34^q+x^2+4^2q-4^3qx+4^4q=671^z$$ but as we know that $(x+4^q)^4>x^4-x^34^q+x^2+4^2q-4^3qx+4^4q$ so case $(V)$ gives $3^{4z}>671^z \implies 3^4>671$ a contradiction So there is one possibility $x=4^q$ which gives $x^5=4^{5q}=4^y$ which implies that $2|2013^z$ again a contradiction. So there do not exists any positive integers $x$, $y$ and $z$ such that $$x^5+4^y=2013^z$$ Is My Solution correct?

Answer 1

Your presentation is rather messy but the arguments at their core are correct.

Your proof would be easier to parse (and check!) if you improved its layout; divide it into a few smaller claims and proofs. For example:

1. Claim: If $(x,y,z)$ is a solution then $5\mid y$.

   Proof. ...

2. Claim: If $(x,y,z)$ is a solution and $y=5q$ then either $$x+4^q=11^z\qquad\text{ or }\qquad x+4^q=3^z.$$

   Proof. ...

3. Etcetera.

As for details:

1. A lot of exponent are typeset incorrectly; you write thing like $3^4z$ in stead of $3^{4z}$ quite a few times.
2. Roman numerals are I, II, III, IV, V, VI, etc. These are upper case i's, not lowercase l's.
3. You can use the environment eqnarray to align several equations, and enumerate them with tag{}. For example: $$\begin{eqnarray} x+4^q&=&11^z, x^4-x^34^q+x^24^{2q}-4^{3q}x+4^{4q}=183^z\tag{I}\\ x+4^q&=&183^z, x^4-x^34^q+x^24^{2q}-4^{3q}x+4^{4q}=11^z\tag{II} \end{eqnarray}$$