Square the nonzero residues modulo eleven. Pairing x with −x avoids duplicate work: 1², 2², 3², 4², 5² give 1, 4, 9, 5, 3. Thus the nonzero quadratic residues are 1, 3, 4, 5, 9. The remaining nonzero classes are nonsquares. This is a statement about existence of an integer square root modulo eleven, not about taking a real square root and rounding it.
Predict
Predict why every nonzero square appears exactly twice among the ten nonzero inputs. If x²≡y², primality implies x≡y or x≡−y. The two are distinct because eleven is odd and y is nonzero. For a composite modulus, a product can vanish without either factor vanishing, so this proof can fail and extra roots can appear.
Experiment
Use a compact square table.
x up to sign
1
2
3
4
5
x² modulo 11
1
4
9
5
3
Locate roots of five and roots of nine. Then try to find a root of two and use the exhaustive table as a certificate for this small prime. A finite table certifies only its specified modulus.
Understand
For an odd prime p, the Legendre symbol (a/p) is zero if p|a, one if a is a nonzero square class and minus one otherwise. There are (p−1)/2 nonzero squares. If g is a primitive root, every unit is g^r, and g^r is square exactly when r is even: squares have even exponents, and an even exponent has the evident square root. Hence (g^r/p)=(−1)^r. Adding exponents proves (ab/p)=(a/p)(b/p), with zero cases handled separately. This proof depends on prime-field cyclicity established earlier, not on an assumed pattern in the table.
Apply
The special class −1 is square precisely when p≡1 modulo four. A root would have order four, so four divides p−1. Conversely a primitive root supplies a fourth-order element when that divisibility holds. This gives an impossibility tool: if an odd prime p≡3 modulo four divides a²+b², then unless both a, b are divisible by p, division by a nonzero square would make −1 a square, impossible. Such arguments will later classify sums of two squares. For a composite modulus, solve at prime-power components and combine by CRT rather than assuming the Legendre definition still applies unchanged.
The two-to-one counting statement excludes zero for a reason. For a nonzero square r modulo an odd prime, its roots x and −x are distinct because two is invertible and x is nonzero. Zero has only the single root zero. Thus there are (p−1)/2 nonzero squares, but (p+1)/2 square residues when zero is included. The Legendre symbol assigns zero to multiples of p, preserving this distinction rather than labeling zero a nonsquare.
Continue
The symbol answers whether a root exists, not necessarily how to compute one efficiently. Primitive-root exponents provide a theoretical construction, but discrete logarithms may be difficult. The next lesson derives Euler’s criterion and Gauss’s lemma, replacing explicit root search with exponentiation or a signed residue count. These mechanisms lead to quadratic reciprocity, which relates residue questions at two different odd primes.
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How many nonzero square classes are there modulo 13?
Integers, fractions and finite decimals are accepted. Expressions such as sqrt(2) are not parsed by this checker.
Find the Legendre symbol (5/11), as −1, 0 or 1.
Integers, fractions and finite decimals are accepted. Expressions such as sqrt(2) are not parsed by this checker.
Prove a nonzero square modulo an odd prime has exactly two roots.
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Use a primitive root to prove multiplicativity of the Legendre symbol.
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Using orders, prove −1 is square modulo odd prime p exactly when p≡1 mod 4.
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Show the two-root statement fails modulo 15.
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Reflect and continue
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