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

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The discussion focuses on finding the minimum sum of integers m and n such that (4^m + 4^n) mod 100 equals 0, with m greater than n. It establishes that for the equation to hold, 4^(m-n) must equal -1 mod 25. The powers of 4 are examined, revealing that 4^5 is congruent to -1 mod 25. Consequently, it is determined that 4^6 + 4^1 equals 0 mod 100, leading to the conclusion that the minimum value of m+n is 7. The solution is confirmed as correct.
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if $(4^m+4^n)$ mod 100=0
(here $m,n\in N \,\, and \,\,m>n$)

please find:$min(m+n)$
 
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[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]
 
Opalg said:
[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 !
 

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