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I am asked if this has one and only one solution. The answer key in my text says that it has 0 or 2 mutually incongruent solutions. How would we know this? I am not asked to find the solutions, but I see manually that the answers are 22 and 78.

HFM
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    $x^2+x = x(x+1)$, so if $a(a+1) \cong 1$, then $-a\cdot -(a+1) \cong 1$ is a distinct solution as long as $-a \not \cong a+1$ (so check, from $2a+1 = 101$ that $a = 50$ and $a+1 = 50$ isn't "accidentally" the solution). – Eric Towers Oct 27 '21 at 04:03
  • The quadratic formula works here - see the linked dupe. – Bill Dubuque Oct 27 '21 at 21:05

2 Answers2

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$101$ is prime, so the integers mod $101$ form a field. A polynomial $P(x)$ over a field has a root $r$ in the field if and only if it is divisible by $x-r$, in which case the quotient $P(x)/(x-r)$ also divides $P(x)$. In particular, if $P(x)$ is a monic quadratic and $x-r$ is a root, the quotient is of the form $x-q$ where $q$ is also a root. The only way $P(x)$ could have only one root in the field is if $q=r$, i.e. $P(x) = (x-r)^2 = x^2 - 2 r x + r^2$. Now if that was $x^2 + x - 1$, we'd need $-2r \equiv 1 \mod 101$, i.e. $r \equiv 50$, but $50^2 \equiv 76 \mod 101$, not $-1$. So $x^2 + x - 1$ doesn't have $1$ root in our field; it must be either $0$ or $2$.

Robert Israel
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$$ 4 x^2 + 4x \equiv 4 \pmod{101} $$

$$ 4 x^2 + 4x + 1 \equiv 5 \pmod{101} $$

$$ (2x+1)^2 \equiv 5 \pmod{101} $$

Well, $5$ is a quadratic residue, there really is a nonzero $w$ such that $ w^2 \equiv 5 \pmod{101} .$ The two square roots are $\pm w.$ Then solve $2x + 1 \equiv w \pmod{101},$ then $2x + 1 \equiv -w \pmod{101}$

Will Jagy
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