Prove that for the DSA:
(α^(e1) β^(e2) mod p) mod q = (α^k mod p) mod q
To prove the given statement, we can use the properties of modular arithmetic, specifically the property that for any integers a, b, c and modulus m:
(a * b) mod m = ((a mod m) * (b mod m)) mod m
(a + b) mod m = ((a mod m) + (b mod m)) mod m
Given equation: (α^(e1) β^(e2) mod p) mod q = (α^k mod p) mod q
Expand the left-hand side:
(α^(e1) β^(e2) mod p) mod q = ((α^(e1) mod p) * (β^(e2) mod p)) mod q
Now, use the properties of modular arithmetic:
((α^(e1) mod p) * (β^(e2) mod p)) mod q = (((α^(e1) mod p) mod q) * ((β^(e2) mod p) mod q)) mod q
Since q is a prime number, it is relatively prime to p, so we can use Fermat's Little Theorem:
α^(p-1) mod p = 1
β^(p-1) mod p = 1
Therefore, α^(e1) mod p = α^(e1 mod (p-1)) mod p = α^(e1 mod (p-1)) mod p
And, β^(e2) mod p = β^(e2 mod (p-1)) mod p = β^(e2 mod (p-1)) mod p
Substitute these values back into the equation:
(((α^(e1 mod (p-1)) mod p) mod q) * ((β^(e2 mod (p-1)) mod p) mod q)) mod q
Since e1 mod (p-1) = k mod (p-1) by definition of DSA:
((((α^k mod p) mod p) mod q) * ((β^(e2 mod (p-1)) mod p) mod q)) mod q
Since β^(e2) mod p = β^(e2 mod (p-1)) mod p, and e2 mod (p-1) = k mod (p-1):
(((α^k mod p) mod p) mod q) * ((β^(e2 mod (p-1)) mod p) mod q)) mod q
Since β^(e2 mod (p-1)) mod q = β^(k mod (p-1)) mod q by definition of DSA:
((α^k mod p) mod q) * (β^(k mod (p-1)) mod q) mod q
Since α^k mod p = α^k mod p:
(α^k mod p) mod q
Therefore, the left-hand side is equal to the right-hand side, and the given statement is proved.