<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] $$a+b\equiv9^n(1+9)\pmod{100}$$
Now $9^n\equiv(-1)^n\pmod{10}\implies10\cdot9^n\equiv10\cdot(-1)^n\pmod{10\cdot10}$
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