A Modular Arithmetic Equation | Number Theory

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SyberMath

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(x+1)² ≡ 2 (mod 7)

a ≡ 0 (mod 7) ⇒ a² ≡ 0 (mod 7)
a ≡ 1 (mod 7) ⇒ a² ≡ 1 (mod 7)
a ≡ 2 (mod 7) ⇒ a² ≡ 4 (mod 7)
a ≡ 3 (mod 7) ⇒ a² ≡ 2 (mod 7)

Then since a ≡ 4, 5, 6, 7 (mod 7) are equivalent to a ≡ -3, -2, -1, 0 (mod 7), so after squaring they will become the same as the negative becomes positive.
So a² ≡ 0, 1, 4, 2 (mod 7)
If a² ≡ 2 (mod 7) then a ≡ 3 (mod 7) or a ≡ 4 (mod 7)
So we must have x+1 ≡ 3 or 4 (mod 7)
Thus x ≡ 2 or 3 (mod 7)

user

Generic approach.

x^2 + 2x + 7n = 1
x^2 + 2x + 7n-1 = 0

We want x to be an integer, so the determinant needs to be a perfect square
(2)^2 - 4(1)(7n-1) = m^2
4 - 4(7n-1) = m^2
4(2 - 7n) = m^2

Note that m will have to be even, since the LHS is even. So, let m = 2k.
4(2-7n) = (2k)^2
2 - 7n = 2k
2 - 2k = 7n
2(1-k) = 7n

We can make this work by letting k = 8. Then n = -2. Going backto the original equation, (7n-1) is now -15, so...

x^2 + 2x - 15 = 0
x^2 + 2x + 1 = 16
x + 1 = +/- 4
x = -1 +/- 4
x = -5 or 3

In the mod 7 world, this is 2 or 3.

chaosredefined

We can used the table and the exercice become not difficulte

prof.mohamad

I don't think it is a universal method, it seems like a coincidence that there are exactly two solutions and then they should correspond to the roots.

miro.s