Prove eachh mathematical statement for integers m,n,k,p,a,b
12. For every prime p and nonzero n we have n^2 is congruent to nMODp if and only if either n is congruent to 1MODp or n is congruent to 0MODp.
To prove the mathematical statement, we will need to show two claims:
Claim 1: If n^2 ≡ n (mod p), then n ≡ 0 (mod ≡ 1 (mod p).
Claim 2: If n ≡ 0 (mod ≡ 1 (mod p), then n^2 ≡ n (mod p).
Let's prove each claim separately:
Claim 1: If n^2 ≡ n (mod p), then n ≡ 0 (mod ≡ 1 (mod p).
To prove Claim 1, we assume that n^2 ≡ n (mod p).
Case 1: n ≡ 0 (mod p)
If n ≡ 0 (mod p), then n can be written as n = px, where x is some integer. Substituting this into the equation n^2 ≡ n (mod p), we get (px)^2 ≡ px (mod p), which simplifies to p^2x^2 ≡ px (mod p). Since p is prime, it is nonzero modulo p. Therefore, we can divide both sides of the equation by p, resulting in px ≡ x (mod p). Since x is an integer, px ≡ x (mod p) implies x ≡ 0 (mod p). Thus, n ≡ 0 (mod p) satisfies the condition.
Case 2: n ≢ 0 (mod p)
If n ≢ 0 (mod p), then n can be written as n = py + r, where y is some integer and r is a nonzero remainder when dividing n by p. Substituting this into the equation n^2 ≡ n (mod p), we get (py + r)^2 ≡ py + r (mod p). Expanding the expression, we have p^2y^2 + 2p yr + r^2 ≡ py + r (mod p). Since we're working modulo p, the first term p^2y^2 is congruent to 0 (mod p^2). Also, since n ≢ 0 (mod p), the second term 2p yr must be congruent to 0 (mod p). Hence, the equation simplifies to r^2 ≡ r (mod p). Now, if r ≡ 0 (mod p), then n ≡ 0 (mod p), which satisfies the condition. If r ≢ 0 (mod p), then we can multiply both sides of the equation by r^(-1), the multiplicative inverse of r modulo p. This gives us r ≡ 1 (mod p), which implies n ≡ 1 (mod p), satisfying the condition.
Therefore, in all cases, if n^2 ≡ n (mod p), then n ≡ 0 (mod ≡ 1 (mod p).
Claim 2: If n ≡ 0 (mod ≡ 1 (mod p), then n^2 ≡ n (mod p).
To prove Claim 2, we assume that n ≡ 0 (mod ≡ 1 (mod p).
Case 1: n ≡ 0 (mod p)
If n ≡ 0 (mod p), then n can be written as n = px, where x is some integer. Substituting this into the equation n^2 ≡ n (mod p), we get (px)^2 ≡ px (mod p), which simplifies to p^2x^2 ≡ px (mod p). This equation can be rewritten as px(p - x) ≡ 0 (mod p). Since p is prime, it is nonzero modulo p. Therefore, we have either px ≡ 0 (mod p) or (p - x) ≡ 0 (mod p). In either case, we conclude that n^2 ≡ n (mod p).
Case 2: n ≡ 1 (mod p)
If n ≡ 1 (mod p), then n can be written as n = py + 1, where y is some integer. Substituting this into the equation n^2 ≡ n (mod p), we get (py + 1)^2 ≡ py + 1 (mod p). Expanding the expression, we have p^2y^2 + 2py + 1 ≡ py + 1 (mod p). Simplifying the equation, we have p^2y^2 ≡ 0 (mod p), which implies that py^2 ≡ 0 (mod p) since p is nonzero modulo p. Consequently, we can rewrite the equation as py(py - 1) ≡ 0 (mod p). Since n ≡ 1 (mod p) implies py ≡ 1 (mod p), we have (py - 1) ≡ 0 (mod p). Therefore, in this case as well, n^2 ≡ n (mod p).
Hence, both claims have been proven, which completes the proof of the mathematical statement.
Therefore, for every prime p and nonzero n, we have n^2 ≡ n (mod p) if and only if either n ≡ 0 (mod ≡ 1 (mod p).