A naive user of RSA has announced the public key $n = 2903239, e = 5$. Eve has worked out that $n = 1237×2347$. Verify that the public key is valid and explain why the private key is $d = 2319725$. [Calculators are not required.]
This question comes from a mock exam that I am trying to solve for tomorrow.
I attach my working out:
$n = p\;q = (1237)(2347) = 2903239$
$\varphi(n) = (1237-1)(2347-1) = 1236(2346) = 2899656$
$d\;e \bmod \varphi(n) = 1$
$e = 5$
$5\;d \bmod 2899656 = 1$
Applying the Euclidean algorithm:
$5\;x + 2899656\;y = 1$
$2899656 = 5(579931)+1$
- The public key part: $(e,n) = (5, 2903239)$
- The private key part: $(d,n) = (2319725,1)$
I do not understand how the private key is calculated, nor how to do this with a calculator. Any help would be welcome.
EDIT: $$\varphi (n) = (1237-1)(2347-1)= 2899656 \neq 2903239$$
de mod (phi(n)) = 1
5d mod 2903239 = 1
5x + 2903239y = 1
2903239 = 5(580647)+4
5 = 4(1) + 1
1 = 5 - 4(1)
1 = 5 - (2903239 - 5(580647)
1 = 5 - 2903239 + 5(580647)
1 = 5(580648) - 1(2903239)
I am still unsure of how to obtain the private key value from this.
2903239 - 5860648 = 2322591
This value is not the same as the desired value of d, d = 2319725.
