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Suppose H(xy) = H(x) * H(y) . Clearly preimage properties are violated. How can we find x, y such that H(x) = H(y) mod (2^k)

Sumish Ajmani
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    x=y would be an obvious solution. – Artjom B. Apr 21 '16 at 22:58
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    I'm voting to close this question as off-topic because it is not a question directly related to programming. – Phiter Apr 21 '16 at 23:01
  • @PhiterFernandes Not all *tags* are directly related to programming. The justification of such tags, and such questions, is that these topics are manifestly *indirectly* related to programming. Computer programming doesn't have sharp boundaries. – John Coleman Apr 21 '16 at 23:04
  • I know, but this is sort of a math thing. – Phiter Apr 21 '16 at 23:07
  • If you define a hash function as a function which maps integers of arbitrary size to integers of a fixed size then the only hash functions that can satisfy your condition are hash functions whose values are always either 0 and 1, in which case your problem is trivially easy to solve. Perhaps you meant something like `H(ab) = H(a)H(b) mod 2^k` where the hash size is k-bits? – John Coleman Apr 21 '16 at 23:29
  • @JohnColeman - ya you are right. I should have written mod 2^k . If you can tell me the answers I would be really obliged to you. – Sumish Ajmani Apr 21 '16 at 23:44
  • @PhiterFernandes - if u know then plzz tell me or email me. I am eager to know this. – Sumish Ajmani Apr 21 '16 at 23:44
  • Sounds like homework. What have you tried? Also -- do you know of any hash functions which this property (other than `H(x) = x (mod 2^k)`)? – John Coleman Apr 21 '16 at 23:47
  • @JohnColeman - Actually the thing is the above problem was given with a hint related to use of Euler totient function ; presently I don't know the hash function, but i am sure you can surely use the above hint. – Sumish Ajmani Apr 22 '16 at 00:05
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    Please see crypto.stackexchange.com for these kinds of questions. Some more background would likely be helpful to separate it from a basic homework question, however. – Rob Napier Apr 22 '16 at 00:37
  • Euler's theorem is probably what the hint is alluding to. – John Coleman Apr 22 '16 at 00:38
  • @JohnColeman : ya, that's what the hint was given. Any proceedings . update ?? I tried a lot using a fact about Euler totient function that x^(phi(2^k)) = 1 mod 2^k . But was unable to proceed – Sumish Ajmani Apr 23 '16 at 02:42

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2^k is a very special modulus. You can prove the following strengthened version of Euler's theorem:

Suppose that x is any postive integer. Then x^phi(2^k) (mod 2^k) is equal to either 0 or 1, with 1 if and only if x is odd.

Proof:

If x is odd, then gcd(x,2^k) = 1, hence x^phi(2^k) (mod 2^k) = 1 by Euler's theorem.

Suppose that x is even. The result is trivially true if x = 0, so suppose that x > 0. Write x = (2^s)*y where y is odd and s > 0. Note that

phi(2^k)` = 2^(k-1)

But then,

x^phi(2^k) = (2^s*y)^phi(2^k)
           = (2^s)^phi(2^k) * y^phi(2^k)
           = 2^(s*phi(2^k)) * y^phi(2^k)
           = 2^(s*2^(k-1)) * y^phi(2^k)
           = 0 * y^phi(2^k) = 0 (mod 2^k)

the last line follows from the fact that s*2^(k-1) >= k hence 2^(s*2^(k-1)) is a multiple of 2^k.

Note that if x is even then you actually have x^k = 0 (mod 2^k), so raising x to the power phi(2^k) is overkill for any but the smallest k.

Given this lemma, it is now trivial to see that there exists distinct x,y with either H(x) = H(y) = 0 or H(x) = H(y) = 1. Since it seems to be homework, I'll leave the details to you.

John Coleman
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