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3^x ≡ -1 (mod 4) implies x ≡ 1 (mod 2)
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If 3\^a ≡ 1 (mod 4) then 3\^(a + 1) ≡ 3 ≡ –1 (mod 4)

If 3\^b ≡ –1 (mod 4) then 3\^(b + 1) ≡ –3 ≡ 1 (mod 4)

So basically the sequence {3\^x (mod 4) : x = 1, 2, 3, ...} is {–1, 1, –1, 1, ...} alternating forever.

So if 3\^x ≡ –1 (mod 4) then it's an odd term of the sequence; x ≡ 1 (mod 2)
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Odd powers of 3 are one less than a multiple of 4.

I.e. 3^(n)+1 can be factored with one factor of (3+1) if n is odd.
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*Suggestion:* Consider the contrapositive.

If it is *not* the case that *x* ≡ 1 (mod 2), then *x* must be even. (I'm implicitly assuming, of course, that *x* is a nonnegative integer. After allf *x* were not an integer, it would be unusual to consider *x* (mod 2); if *x* were not a *nonnegative* integer, then 3^(*x*) would not be an integer, in which case discussing 3^(*x*) (mod 4) would likewise be curious.) Thus *x* = 2*y*, *y* another nonnegative integer, in which case

- 3^(*x*)

 = 3^(2*y*)

 = (3^(2))^(*x*)

 = 9^(*y*). **(1)**

What happens when we reduce **(1)** modulo 4? Is the result compatible with having 3^(*x*) ≡ -1 (mod 4)?

Hope this helps. Good luck!
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What does 3^x + 1 = 4m for some integer m tell us about 3^x?

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