Prove that for the DSA:
(αe1 βe2 mod p) mod q = (αk mod p) mod q
To prove this, we need to show that:
(αe1 βe2 mod p) mod q = (αk mod p) mod q
First, let's expand the left side of the equation:
(αe1 βe2 mod p) mod q = ((α^e1 mod p) (β^e2 mod p)) mod q
Since e1 = k and e2 = 1 (based on the definition of the DSA), we can rewrite this as:
((α^k mod p) (β mod p)) mod q
Since β is usually a constant in the DSA algorithm (g^x mod p), we can simplify this further:
(α^k (g^x mod p)) mod q
Now let's expand the right side of the equation:
(αk mod p) mod q
This can be simplified to:
(α^k mod p) mod q
Therefore, we have shown that:
(αe1 βe2 mod p) mod q = (αk mod p) mod q
Hence, we have successfully proved that for the DSA, the equation holds true.