<s>[INST] If $n\geq 3$, $9^n \equiv a (\mod 100)$ and $9^{n+1} \equiv b (\mod 100)$, then $a+b=90$. I noticed a pattern in the powers of 9 modulo 100.
$9^1 \equiv 9 \pmod{100}$
$9^2 \equiv 81 \pmod{100}$
$9^3 \equiv 29 \pmod{100}$
$9^4 \equiv 61 \pmod{100}$
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and conjectured the following:
If $n\geq 3$ is an odd integer where $9^n \equiv a \pmod{100}$ and $9^{n+1} \equiv  b \pmod{100}$, then $a+b=90$.
How do I prove this?
 [/INST] From the binomial theorem, it follows that, 
for $n$ odd, $9^n=(10-1)^n\equiv10n-1\bmod100$, 
and  $9^{n+1}=(10-1)^{n+1}\equiv-10(n+1)+1\bmod100$.  
Therefore, for $n$ odd, $9^n+9^{n+1}\equiv-10\equiv90\bmod100$.
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