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Assume d is a 128 bit random integer and P is base point of an elliptic curve and Q = dP is a point on the elliptic curve and SHA is a hash function with 128 bit output, my question is:
Is size Q equal to size SHA(Q)?
If not, then wich is smaller size?

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    $\begingroup$ Duplicate of math.stackexchange.com/questions/475049/… $\endgroup$ – minar Aug 24 '13 at 16:01
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    $\begingroup$ "The elliptic curve" is undefined, and very relevant. "Size (of) Q" can have several interpretations, for there are several means to define a point on an elliptic curve. No hash form the SHA family is 128-bit. $\endgroup$ – fgrieu Aug 24 '13 at 16:56
  • $\begingroup$ Suppose we take only 128-bits out of 160-bits output in SHA-1 hash function. $\endgroup$ – star Aug 24 '13 at 17:10
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    $\begingroup$ @star Consider this: what if we replace "elliptic curve" in your question with a different finite field, for example the ring of integers modulo n. Do you still think your question can be answered? Don't you think we'd need more information, for example the modulus? $\endgroup$ – orlp Aug 24 '13 at 18:30
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This was answered in the Comments of the math.stackex:

Recall SHA-1 produces a 'fixed length' output, regardless of input, of a 160-bits. For elliptic curves, we could have the points (0,0) all the way up to two 128-bit points (P,Q) if that is your underlying field size. So, you need to be able to handle points that can be as large as 128-bits, but those can be representing integers equal to 0 all the way up to 128-bits. In other words, Q can be zero, but SHA(Q) is 160-bits. Also, I suppose you meant Q to be one of the points (x,y) on the elliptic curve. – Amzoti

Meaning no, they're not equal SHA-1 is larger.

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