h0pe
  • h0pe
Find an integer x such that 0 <= x < 527 and x^37 ≡3 (mod 527).
Mathematics
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h0pe
  • h0pe
Find an integer x such that 0 <= x < 527 and x^37 ≡3 (mod 527).
Mathematics
schrodinger
  • schrodinger
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UsukiDoll
  • UsukiDoll
so we need a number x such that the remainder will be 3 in mod 527. Trial and error will take too long because the exponent is just too big here
h0pe
  • h0pe
Then what should I try?
ganeshie8
  • ganeshie8
start by factoring 527

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h0pe
  • h0pe
17 and 31
ganeshie8
  • ganeshie8
that means \(\phi(527) = (17-1)(31-1) = 480\)
ganeshie8
  • ganeshie8
Now solve below equation \[37y \equiv 1 \pmod{480}\]
h0pe
  • h0pe
I have to find the inverse of 37, right?
ganeshie8
  • ganeshie8
that is, find the inverse of exponent "37" in mod 480
ganeshie8
  • ganeshie8
yes
h0pe
  • h0pe
It's 13
ganeshie8
  • ganeshie8
\[\large x \equiv 3^{13} \pmod{527}\]
h0pe
  • h0pe
\[x\equiv1594323(\mod527)\] which is \[x\equiv148 (\mod527)\]?
ganeshie8
  • ganeshie8
Looks good!
ganeshie8
  • ganeshie8
since they want x to be between 0 and 527, just pick x = 148
h0pe
  • h0pe
thank you!
ganeshie8
  • ganeshie8
as you can see breaking RSA is trivial when you're able to factor the modulus
h0pe
  • h0pe
That was RSA?
ganeshie8
  • ganeshie8
RSA is the name of an encryption system, I thought you're working on encryption algorithms... I went through your past questions...
h0pe
  • h0pe
Oh, so that's what it's called..

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