Solve x²+2x−2≡0 modulo eleven. Completing the square gives (x+1)²≡3. The square table shows roots of three are five and six, so x≡4 or 5. Substitution verifies both. This method turns a quadratic equation into a square-residue question and then returns to the original variable, rather than stopping after announcing that a discriminant is a square.
Predict
Predict how many roots a nondegenerate quadratic can have over a prime field. The discriminant may be a nonsquare, zero or a nonzero square, giving zero, one or two roots respectively. Modulo a composite integer the count can be larger because independent choices at different prime factors combine. A quadratic’s degree does not bound its number of roots over a ring with zero divisors.
Experiment
Combine roots of one modulo fifteen.
Root modulo 3
Root modulo 5
Root modulo 15
1
1
1
1
−1
4
−1
1
11
−1
−1
14
Verify every CRT representative and its square. The component lists are exhaustive, so the four rows prove completeness rather than merely exhibit four examples.
Understand
For an odd prime p and p∤a, multiplication by 4a gives 4a(ax²+bx+c)=(2ax+b)²−(b²−4ac). Since 4a and 2a are invertible, solving the quadratic is equivalent to finding square roots y of the discriminant and setting x=(y−b)(2a)⁻¹. If a vanishes modulo p, solve the remaining linear or constant equation separately. If p=2, enumerate its two classes or use a separate binary analysis; dividing by two is invalid. The prime-field theorem cleanly classifies roots only when these hypotheses are respected.
Apply
For an odd positive composite n with factorization ∏pᵢ^eᵢ, define the Jacobi symbol (a/n)=∏(a/pᵢ)^eᵢ. It is multiplicative, and reciprocity-style rules extend to it, but its value one need not mean a is square modulo n. For example, two is nonsquare modulo both three and five, yet (2/15)=1. An actual congruence solution must exist at every prime-power component. Solve those component equations and use CRT to combine all choices. Distinct component tuples give distinct classes modulo n. Prime powers require lifting analysis rather than simply repeating the prime roots unchanged.
For a composite modulus, even the number of roots carries information. With distinct odd primes p, q and a target that is a nonzero square modulo both, each component has two roots and CRT combines their independent choices into four roots modulo pq. If the target is zero in one component, that component supplies only one root. Prime powers with singular roots can behave differently again, which is why the next lifting lesson tracks derivatives and compatibility at each stage instead of extrapolating the field root bound.
Continue
A root-finding report should contain the reduction, each component solution list, the CRT combinations and direct substitution. A single root does not prove completeness, and a positive Jacobi symbol is only partial evidence for composite moduli. Next, Hensel lifting determines which roots modulo p extend to roots modulo p² and beyond, distinguishing the simple-derivative case from singular roots that may branch or disappear.
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How many classes modulo 15 solve x²≡1?
Integers, fractions and finite decimals are accepted. Expressions such as sqrt(2) are not parsed by this checker.
Find the Jacobi symbol (2/15).
Integers, fractions and finite decimals are accepted. Expressions such as sqrt(2) are not parsed by this checker.
For odd prime p with p∤a, prove ax²+bx+c≡0 is equivalent to (2ax+b)²≡b²−4ac.
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Solve x²−3x+2≡0 modulo 7.
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List all roots of x²≡1 modulo 21.
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Explain why Jacobi (2/15)=1 does not mean 2 is a square modulo 15.
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Reflect and continue
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