Smallest Prime-Factor of $4^{52} + 52^{2013} + 2013^{52}$

Question

I would like to find out the smallest prime factor of $4^{52} + 52^{2013} + 2013^{52}$ by hand. Thanks in advance

Answer 1

The given number is odd, so it cannot be divided by $2$.

In mod $3$, we have $$\begin{align}4^{52}+52^{2013}+2013^{52}\equiv 1+1+0\equiv 2\end{align}$$

In mod $5$, we have $$\begin{align}4^{52}+52^{2013}+2013^{52}&\equiv (-1)^{52}+2^{2013}+3^{52}\\&\equiv 1+2\cdot 2^{2012}+(-1)^{26}\\&\equiv 1+2\cdot {16}^{503}+1\\&\equiv 1+2+1\\&\equiv 4\end{align}$$

In mod $7$, we have $$\begin{align}4^{52}+52^{2013}+2013^{52}&\equiv (-3)^{52}+3^{2013}+4^{52}\\&\equiv \color{red}{2^{26}}+3\cdot 2^{1006}+\color{red}{2^{26}}\\&\equiv \color{red}{2^{27}}+3\cdot 4^{503}\\&\equiv 2\cdot 4^{13}+3\cdot 4\cdot {2}^{251}\\&\equiv 2\cdot 4\cdot {2}^6+5\cdot 2^{251}\\& \equiv 1\cdot 2^6+5\cdot 2\cdot 4^{125}\\&\equiv 64+3\cdot 4\cdot 2^{62}\\&\equiv 1+5\cdot 4^{31}\\&\equiv 1+5\cdot 4\cdot 2^{15}\\&\equiv 1+6\cdot 2\cdot {4}^{7}\\&\equiv 1+5\cdot 4\cdot 2^{3}\\&\equiv 1+6\cdot 1\\&\equiv 0 \end{align}$$ Hence, the answer is $7$.

Answer 2

Essentially, this is the same idea as mathlove's answer, but using Fermat's Little Theorem, we can simplify the computation a bit.

Using modular multiplication, we can reduce the base mod $p$. Using Fermat's Little Theorem, as long as he base is not $0$, we can reduce the exponent mod $p-1$.

$4^{52}\equiv0\pmod{2}$
$52^{2013}\equiv0\pmod{2}$
$2013^{52}\equiv1^0\equiv1\pmod{2}$
$4^{52}+52^{2013}+2013^{52}\equiv1\pmod{2}$

$4^{52}\equiv1^{0}\equiv1\pmod{3}$
$52^{2013}\equiv1^{1}\equiv1\pmod{3}$
$2013^{52}\equiv0\pmod{3}$
$4^{52}+52^{2013}+2013^{52}\equiv2\pmod{3}$

$4^{52}\equiv4^0\equiv1\pmod{5}$
$52^{2013}\equiv2^1\equiv2\pmod{5}$
$2013^{52}\equiv3^0\equiv1\pmod{5}$
$4^{52}+52^{2013}+2013^{52}\equiv4\pmod{5}$

$4^{52}\equiv4^4\equiv4\pmod{7}$
$52^{2013}\equiv3^3\equiv6\pmod{7}$
$2013^{52}\equiv4^4\equiv4\pmod{7}$
$4^{52}+52^{2013}+2013^{52}\equiv14\equiv0\pmod{7}$

Answer 3

As I've commented, the divisibility of $4^{52} + 52^{2013} + 2013^{52}$ by $7$ can be derived very easily as follows

$\displaystyle52\equiv3\pmod7\implies52^{2013}\equiv3^{2013}$

Now as $\displaystyle3^3=27\equiv-1\pmod7$ and $\displaystyle2013=3\cdot\text{ some odd number (namely}, 671 ),$

$\displaystyle 3^{2013}\equiv(-1)^{\text{ some odd number }}\pmod7\equiv-1\ \ \ \ (1)$

Again as $2013\equiv4\pmod 7\implies 2013^{52}\equiv4^{52},$

$\displaystyle 4^{52}+2013^{52}\equiv2\cdot4^{52}\pmod7\equiv2\cdot2^{104}=2^{105}=(2^3)^{35}\equiv1^{35}\pmod7$ as $2^3=8\equiv1\pmod7$

$\displaystyle\implies4^{52}+2013^{52}\equiv1\pmod7\ \ \ \ (2)$