rsa key generation

[zxcvbnm,.]

I am trying to create a program for the RSA algorithm. I have managed to get the public keys, but I hit a brick wall trying to figure out how to generate the secret key. I have tried, but the output is either incorrect or doesn't output at all. my program for the secret key is

d=1;
while (d!=(1/e)%phi)
{
     d++;
}

where d is the secret key, e is the relatively prime number to phi, phi being (p-1)*(q-1), where p and q are the two primes.

can anyone tell me what's wrong?

[Moschops]

Is e an integer?

What do you think 1/e will give you?

[zxcvbnm,.]

ah, I see. but even so, how would I implement this?

d=1;
for (int i=1; d!=((i*phi+1)/e)%phi;i++)
{
     d++
}

???

[Moschops]

You just did (although missing a semi-colon). What does that not do that you think it should do?

[zxcvbnm,.]

When I insert this into the rest of my algorithm and run the program, the result is this:

First prime: 5
Second Prime: 7
the public key is [17,35]
The secret key is 1
phi is 24.

I used 5 and 7 just for simplicity. here the public key is in the form [e,m], where m is p*q, e the same as above, and the secret key is d. One can see that d doesn't solve the equation e*x-phi*y=1, where y is an arbitrary number assigned "i" in the above code.

17*1-24*y=1
17-24*y=1
-24*y=-16
3*y=2
y=2/3
which is not an integer solution.

[Moschops]

You need to find the value of d such that d*e (mod 24) == 1

int d=1;
for ( ; ((d*e)%24) != 1; d++)
{
  
}

This will start with d equal to one, and will begin by calculating:
1*17 mod 24, which is 17, which is not equal to one, so the loop continues...
2*17 mod 24, which is 10, which is not equal to one, so the loop continues...
...
17*17 mod 24, which is 1, which is equal to one, so the loop stops with d set to 17.

This variable "i" you created is not needed.

[zxcvbnm,.]

Ah so that's how you do it. I had the equation, but i didn't know how to solve for d. thank you!