Question

By inspection, find all least residues r modulo 7 that satisfy r 2≡−5

Answer 1

The least residues ( r ) modulo 7 that satisfy
`\( r^2 \equiv -5 \)` are
`\( r \equiv 3 \) and \( r \equiv 4 \).`

Step-by-step explanation:

To find the least residues ( r ) modulo 7 satisfying
`\( r^2 \equiv -5 \),` we inspect the quadratic residues for each residue class modulo 7. Squaring each residue class from 0 to 6 and taking the remainders modulo 7, we look for residues that are congruent to -5 (mod 7).

`\[0^2 \equiv 0, \quad 1^2 \equiv 1, \quad 2^2 \equiv 4, \quad 3^2 \equiv 2, \quad 4^2 \equiv 2, \quad 5^2 \equiv 4, \quad 6^2 \equiv 1 \pmod{7}\]`

By inspection, we find that
`\(3^2 \equiv 2\)` and
`\(4^2 \equiv 2\)` are the residues satisfying
`\( r^2 \equiv -5 \) (mod 7).` Therefore,
`\( r \equiv 3 \) and \( r \equiv 4 \)` are the least residues modulo 7 that satisfy the given congruence.

This result aligns with the concept of quadratic residues and congruences modulo a prime. The solutions are found by examining the squares of residues and identifying those equivalent to -5 modulo 7. The least residues are chosen to ensure we obtain the smallest positive integers that satisfy the congruence.