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Given E: (elliptical curve)

$y^2 = x_3+2x+2 \bmod 17$

Recall: $y^2 = x_3+ax+b$$y^2 = x^3+ax+b$

point $P=(5,1)$

Compute:

$2P = P+P = (5,1)+(5,1)= (x_3,y_3)$

Now the formula used here is slope $m = \dfrac{3x_1^2 +a}{2y_1}$

and what they got is $s = (2 · 1)^{−1} (3 · 5^2 + 2) = 2^{−1} · 9 ≡ 9 · 9 ≡ 13 \bmod 17$

this is what confuses me.

should it not be $(2·1)^{-1}(3 · 5^2 + 2)= 38.5 \bmod 17$ which is 4.5?

Sorry if this may seem trivial, I'm doing a research for a high school project.

The next steps are:

$x_3 = m^2 − x_1 − x_2 = 13^2 − 5 − 5 = 159 ≡ 6 \bmod 17$ $y_3 = m(x_1−x_3) − y_1 = 13(5 − 6) − 1= −14 ≡ 3 \bmod 17$

$(x_3,y_3) = (6,3)$

which I don't understand as well

If someone could enlighten me on this I would be very happy!

Given E: (elliptical curve)

$y^2 = x_3+2x+2 \bmod 17$

Recall: $y^2 = x_3+ax+b$

point $P=(5,1)$

Compute:

$2P = P+P = (5,1)+(5,1)= (x_3,y_3)$

Now the formula used here is slope $m = \dfrac{3x_1^2 +a}{2y_1}$

and what they got is $s = (2 · 1)^{−1} (3 · 5^2 + 2) = 2^{−1} · 9 ≡ 9 · 9 ≡ 13 \bmod 17$

this is what confuses me.

should it not be $(2·1)^{-1}(3 · 5^2 + 2)= 38.5 \bmod 17$ which is 4.5?

Sorry if this may seem trivial, I'm doing a research for a high school project.

The next steps are:

$x_3 = m^2 − x_1 − x_2 = 13^2 − 5 − 5 = 159 ≡ 6 \bmod 17$ $y_3 = m(x_1−x_3) − y_1 = 13(5 − 6) − 1= −14 ≡ 3 \bmod 17$

$(x_3,y_3) = (6,3)$

which I don't understand as well

If someone could enlighten me on this I would be very happy!

Given E: (elliptical curve)

$y^2 = x_3+2x+2 \bmod 17$

Recall: $y^2 = x^3+ax+b$

point $P=(5,1)$

Compute:

$2P = P+P = (5,1)+(5,1)= (x_3,y_3)$

Now the formula used here is slope $m = \dfrac{3x_1^2 +a}{2y_1}$

and what they got is $s = (2 · 1)^{−1} (3 · 5^2 + 2) = 2^{−1} · 9 ≡ 9 · 9 ≡ 13 \bmod 17$

this is what confuses me.

should it not be $(2·1)^{-1}(3 · 5^2 + 2)= 38.5 \bmod 17$ which is 4.5?

Sorry if this may seem trivial, I'm doing a research for a high school project.

The next steps are:

$x_3 = m^2 − x_1 − x_2 = 13^2 − 5 − 5 = 159 ≡ 6 \bmod 17$ $y_3 = m(x_1−x_3) − y_1 = 13(5 − 6) − 1= −14 ≡ 3 \bmod 17$

$(x_3,y_3) = (6,3)$

which I don't understand as well

If someone could enlighten me on this I would be very happy!

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user3170899
user3170899
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Given E: (elliptical curve)

$y^2 = x_3+2x+2 \bmod 17$

Recall: $y^2 = x_3+ax+b$

point $P=(5,1)$

Compute:

$2P = P+P = (5,1)+(5,1)= (x_3,y_3)$

Now the formula used here is slope $m = \dfrac{3x_1^2 +a}{2y_1}$

and what they got is $s = (2 · 1)^{−1} (3 · 5^2 + 2) = 2−1 · 9 ≡ 9 · 9 ≡ 13 \bmod 17$$s = (2 · 1)^{−1} (3 · 5^2 + 2) = 2^{−1} · 9 ≡ 9 · 9 ≡ 13 \bmod 17$

this is what confuses me.

should it not be $(2·1)^{-1}(3 · 5^2 + 2)= 38.5 \bmod 17$ which is 4.5?

Sorry if this may seem trivial, I'm doing a research for a high school project.

The next steps are:

$x_3 = m^2 − x_1 − x_2 = 13^2 − 5 − 5 = 159 ≡ 6 \bmod 17$ $y_3 = m(x_1−x_3) − y_1 = 13(5 − 6) − 1= −14 ≡ 3 \bmod 17$

$(x_3,y_3) = (6,3)$

which I don't understand as well

If someone could enlighten me on this I would be very happy!

Given E: (elliptical curve)

$y^2 = x_3+2x+2 \bmod 17$

Recall: $y^2 = x_3+ax+b$

point $P=(5,1)$

Compute:

$2P = P+P = (5,1)+(5,1)= (x_3,y_3)$

Now the formula used here is slope $m = \dfrac{3x_1^2 +a}{2y_1}$

and what they got is $s = (2 · 1)^{−1} (3 · 5^2 + 2) = 2−1 · 9 ≡ 9 · 9 ≡ 13 \bmod 17$

this is what confuses me.

should it not be $(2·1)^{-1}(3 · 5^2 + 2)= 38.5 \bmod 17$ which is 4.5?

Sorry if this may seem trivial, I'm doing a research for a high school project.

The next steps are:

$x_3 = m^2 − x_1 − x_2 = 13^2 − 5 − 5 = 159 ≡ 6 \bmod 17$ $y_3 = m(x_1−x_3) − y_1 = 13(5 − 6) − 1= −14 ≡ 3 \bmod 17$

$(x_3,y_3) = (6,3)$

which I don't understand as well

If someone could enlighten me on this I would be very happy!

Given E: (elliptical curve)

$y^2 = x_3+2x+2 \bmod 17$

Recall: $y^2 = x_3+ax+b$

point $P=(5,1)$

Compute:

$2P = P+P = (5,1)+(5,1)= (x_3,y_3)$

Now the formula used here is slope $m = \dfrac{3x_1^2 +a}{2y_1}$

and what they got is $s = (2 · 1)^{−1} (3 · 5^2 + 2) = 2^{−1} · 9 ≡ 9 · 9 ≡ 13 \bmod 17$

this is what confuses me.

should it not be $(2·1)^{-1}(3 · 5^2 + 2)= 38.5 \bmod 17$ which is 4.5?

Sorry if this may seem trivial, I'm doing a research for a high school project.

The next steps are:

$x_3 = m^2 − x_1 − x_2 = 13^2 − 5 − 5 = 159 ≡ 6 \bmod 17$ $y_3 = m(x_1−x_3) − y_1 = 13(5 − 6) − 1= −14 ≡ 3 \bmod 17$

$(x_3,y_3) = (6,3)$

which I don't understand as well

If someone could enlighten me on this I would be very happy!

deleted 3 characters in body
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Michael Hardy
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Given E: (elliptical curve)

$y^{2} = x_3+2x+2$ $mod$ $17$$y^2 = x_3+2x+2 \bmod 17$

Recall: $y^{2} = x_3+ax+b$$y^2 = x_3+ax+b$

point P=(5,1)$P=(5,1)$

Compute:

$2P = P+P = (5,1)+(5,1)= (x_3,y_3)$

Now the formula used here is slope $m = \frac{3x_1^2 +a}{2y_1}$$m = \dfrac{3x_1^2 +a}{2y_1}$

and what they got is $s = (2 · 1)^{−1} (3 · 5^2 + 2) = 2−1 · 9 ≡ 9 · 9 ≡ 13 mod 17$$s = (2 · 1)^{−1} (3 · 5^2 + 2) = 2−1 · 9 ≡ 9 · 9 ≡ 13 \bmod 17$

this is what confuses me.

should it not be $(2·1)^{-1}(3 · 5^2 + 2)= 38.5 mod 17$$(2·1)^{-1}(3 · 5^2 + 2)= 38.5 \bmod 17$ which is 4.5?

Sorry if this may seem trivial, I'm doing a research for a high school project.

The next steps are:

$x_3 = m^2 − x_1 − x_2 = 13^2 − 5 − 5 = 159 ≡ 6 mod 17$$x_3 = m^2 − x_1 − x_2 = 13^2 − 5 − 5 = 159 ≡ 6 \bmod 17$ $y_3 = m(x_1−x_3) − y_1 = 13(5 − 6) − 1= −14 ≡ 3 mod 17$$y_3 = m(x_1−x_3) − y_1 = 13(5 − 6) − 1= −14 ≡ 3 \bmod 17$

$(x_3,y_3) = (6,3)$

which I don't understand as well

If someone could enlighten me on this I would be very happy!

Given E: (elliptical curve)

$y^{2} = x_3+2x+2$ $mod$ $17$

Recall: $y^{2} = x_3+ax+b$

point P=(5,1)

Compute:

$2P = P+P = (5,1)+(5,1)= (x_3,y_3)$

Now the formula used here is slope $m = \frac{3x_1^2 +a}{2y_1}$

and what they got is $s = (2 · 1)^{−1} (3 · 5^2 + 2) = 2−1 · 9 ≡ 9 · 9 ≡ 13 mod 17$

this is what confuses me.

should it not be $(2·1)^{-1}(3 · 5^2 + 2)= 38.5 mod 17$ which is 4.5?

Sorry if this may seem trivial, I'm doing a research for a high school project.

The next steps are:

$x_3 = m^2 − x_1 − x_2 = 13^2 − 5 − 5 = 159 ≡ 6 mod 17$ $y_3 = m(x_1−x_3) − y_1 = 13(5 − 6) − 1= −14 ≡ 3 mod 17$

$(x_3,y_3) = (6,3)$

which I don't understand as well

If someone could enlighten me on this I would be very happy!

Given E: (elliptical curve)

$y^2 = x_3+2x+2 \bmod 17$

Recall: $y^2 = x_3+ax+b$

point $P=(5,1)$

Compute:

$2P = P+P = (5,1)+(5,1)= (x_3,y_3)$

Now the formula used here is slope $m = \dfrac{3x_1^2 +a}{2y_1}$

and what they got is $s = (2 · 1)^{−1} (3 · 5^2 + 2) = 2−1 · 9 ≡ 9 · 9 ≡ 13 \bmod 17$

this is what confuses me.

should it not be $(2·1)^{-1}(3 · 5^2 + 2)= 38.5 \bmod 17$ which is 4.5?

Sorry if this may seem trivial, I'm doing a research for a high school project.

The next steps are:

$x_3 = m^2 − x_1 − x_2 = 13^2 − 5 − 5 = 159 ≡ 6 \bmod 17$ $y_3 = m(x_1−x_3) − y_1 = 13(5 − 6) − 1= −14 ≡ 3 \bmod 17$

$(x_3,y_3) = (6,3)$

which I don't understand as well

If someone could enlighten me on this I would be very happy!

deleted 27 characters in body
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qwr
qwr
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