Find the Min Sum of $m$ and $n$ for $(4^m+4^n)\ mod\ 100=0$

[Albert1]

if $(4^m+4^n)$ mod 100=0
(here $m,n\in N \,\, and \,\,m>n$)

please find:$min(m+n)$

 

[Opalg]

[sp]$4^m+4^n = 4^n(4^{m-n}+1)$. If $4^{m-n} = -1\pmod{25}$ then $4^n(4^{m-n}+1) = 0 \pmod{100}.$ The first few powers of $4$ are

$4$
$16$
$64$
$256$
$1024 = 4^5$.​

Since $4^5 = 24 = -1\pmod{25}$, it follows that $4^6 + 4^1 = 0\pmod{100}$ (and in fact $4^6+4^1 = 4096 + 4 = 4100$). So the minimum value of $m+n$ is $6+1=7.$[/sp]

 

[Albert1]

[sp]$4^m+4^n = 4^n(4^{m-n}+1)$. If $4^{m-n} = -1\pmod{25}$ then $4^n(4^{m-n}+1) = 0 \pmod{100}.$ The first few powers of $4$ are

$4$
$16$
$64$
$256$
$1024 = 4^5$.​

Since $4^5 = 24 = -1\pmod{25}$, it follows that $4^6 + 4^1 = 0\pmod{100}$ (and in fact $4^6+4^1 = 4096 + 4 = 4100$). So the minimum value of $m+n$ is $6+1=7.$[/sp]

very good , the answer is correct !