Simple Encrypting System

x is the Number you want to encrypt. y is the key. r is a random number between 0 and y-1. z is the encryptet number.

z = x*y+r

To decrypt you can use this: x = (z-(z mod y))/y

Is this System safe or can you decrypt z without the key?

Note by Oliver Schnell
3 years ago

No vote yet
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Comments

Without the key, it doesn't seem very crackable from a linear algebra point of view (we could make linear equations by taking logarithms). Consider the simpler case with r=0r = 0 . Let's say we had several samples:

z1=x1 yz2=x2 yz3=x3 yz_1 = x_1 \, y \\ z_2 = x_2 \, y \\ z_3 = x_3 \, y

Here, there are three equations and four unknowns. Adding more equations always results in an under-defined system. If we consider the r≠0r \neq 0 case, it gets even worse, because we add two new unknowns for each new equation. If many samples are taken, we can perhaps infer something about y from the sum of all of the r values (given assumptions about the distribution of r), resulting in one extra equation. But this comes nowhere near to balancing the liability incurred from introducing all the new unknown r values.

If I'm wrong about this, I'd be fascinated to hear why / how

Steven Chase - 3 years ago
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