As in the title, I'm trying to prove $-12^{482}$ has a multiplicative inverse modulo 175. Generally, I believe I could do this by showing that $gcd(-12^{482}, 175)=1$, (i.e that they're relatively prime) but I'm not sure about how to deal with the exponent without fully computing $-12^{482}$.
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1Do you think that $2^{100}$ is relatively prime to $5$? – B. Goddard Mar 06 '21 at 13:33
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Yes. That helps. – Jonathan Sudelko Mar 06 '21 at 13:41
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More explicitly, since $$12 = 2^2 \times 3^1,$$ $$12^{(482)} = 2^{(482) \times 2} \times 3^{(482) \times 1}.$$ Thus, the only prime factors in $12^{(482)}$ are $2$ and $3$. Both $2$ and $3$ are relatively prime to $(175) = 5^2 \times 7.$ – user2661923 Mar 06 '21 at 13:42
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Yes, $a$ is invertible mod $n\iff a$ is coprime to $n$, and coprimes to $n$ are closed under product by the linked dupe, so it suffices to show that $12$ (and $-1$) are coprime to $175\ \ $ – Bill Dubuque Mar 06 '21 at 15:52
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Hint: $x^k$ is invertible modulo $n$ iff $x$ is. This is because $x$ and $x^k$ have the same prime factors, just with different multiplicities.
In your case, $n = 175 = 5^2 \cdot 7$ while $x = 12 = 2^2 \cdot 3$ are coprime.
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