How to find x-coordinate of the curve If y-coordinate is given?

disp('y^2 = x^3 + 14 mod 17')
y=0:16
for i=1:17
    left_side = mod(y(i).^2,17);
    right_side = mod(x.^3+14,17;
end

If I have y-coordinates from 0 to 16, how can I find the corresponding x-coordinate that satifies 'y^2 = x^3 + 14 mod 17'

Answer 1

[Image Analyst]

I understand now. Try this:

for k1 = 0 : 16
	for k2 = 0 : 16
		equation1 = k1 .^ 2;
		equation2 = k2 .^ 3 + 14;
		eqn1mod = mod(equation1, 17);
		eqn2mod = mod(equation2, 17);
		indexes = find(eqn1mod == eqn2mod);
		if ~isempty(indexes)
			fprintf('Match at k1 = %d and k2 = %d.\n', k1, k2);
			fprintf('    equation1 = %d and equation2 = %d.\n', equation1, equation2);
			fprintf('    mod(equation1, 17) = %d and mod(equation2, 17) = %d.\n', eqn1mod, eqn2mod);
		end
	end
end

You'll see

Match at k1 = 0 and k2 = 7.

equation1 = 0 and equation2 = 357.

mod(equation1, 17) = 0 and mod(equation2, 17) = 0.

Match at k1 = 1 and k2 = 13.

equation1 = 1 and equation2 = 2211.

mod(equation1, 17) = 1 and mod(equation2, 17) = 1.

Match at k1 = 2 and k2 = 14.

equation1 = 4 and equation2 = 2758.

mod(equation1, 17) = 4 and mod(equation2, 17) = 4.

Match at k1 = 3 and k2 = 6.

equation1 = 9 and equation2 = 230.

mod(equation1, 17) = 9 and mod(equation2, 17) = 9.

Match at k1 = 4 and k2 = 8.

equation1 = 16 and equation2 = 526.

mod(equation1, 17) = 16 and mod(equation2, 17) = 16.

Match at k1 = 5 and k2 = 12.

equation1 = 25 and equation2 = 1742.

mod(equation1, 17) = 8 and mod(equation2, 17) = 8.

Match at k1 = 6 and k2 = 11.

equation1 = 36 and equation2 = 1345.

mod(equation1, 17) = 2 and mod(equation2, 17) = 2.

Match at k1 = 7 and k2 = 1.

equation1 = 49 and equation2 = 15.

mod(equation1, 17) = 15 and mod(equation2, 17) = 15.

Match at k1 = 8 and k2 = 16.

equation1 = 64 and equation2 = 4110.

mod(equation1, 17) = 13 and mod(equation2, 17) = 13.

Match at k1 = 9 and k2 = 16.

equation1 = 81 and equation2 = 4110.

mod(equation1, 17) = 13 and mod(equation2, 17) = 13.

Match at k1 = 10 and k2 = 1.

equation1 = 100 and equation2 = 15.

mod(equation1, 17) = 15 and mod(equation2, 17) = 15.

Match at k1 = 11 and k2 = 11.

equation1 = 121 and equation2 = 1345.

mod(equation1, 17) = 2 and mod(equation2, 17) = 2.

Match at k1 = 12 and k2 = 12.

equation1 = 144 and equation2 = 1742.

mod(equation1, 17) = 8 and mod(equation2, 17) = 8.

Match at k1 = 13 and k2 = 8.

equation1 = 169 and equation2 = 526.

mod(equation1, 17) = 16 and mod(equation2, 17) = 16.

Match at k1 = 14 and k2 = 6.

equation1 = 196 and equation2 = 230.

mod(equation1, 17) = 9 and mod(equation2, 17) = 9.

Match at k1 = 15 and k2 = 14.

equation1 = 225 and equation2 = 2758.

mod(equation1, 17) = 4 and mod(equation2, 17) = 4.

Match at k1 = 16 and k2 = 13.

equation1 = 256 and equation2 = 2211.

mod(equation1, 17) = 1 and mod(equation2, 17) = 1.

[Image Analyst]

Sure, go ahead if you want to.

[Image Analyst]

Just set up an array and index and assign them. You'll need to learn how to assign values into a matrix if you're going to be successful in MATLAB. You can start here:

MATLAB Academy - Free 2 hour training

In the meantime, here is how you can store the values in an array called "results".

results = [0, 0];
counter = 1;
for k1 = 0 : 16
	for k2 = 0 : 16
		equation1 = k1 .^ 2;
		equation2 = k2 .^ 3 + 14;
		eqn1mod = mod(equation1, 17);
		eqn2mod = mod(equation2, 17);
		indexes = find(eqn1mod == eqn2mod);
		if ~isempty(indexes)
			fprintf('Match at k1 = %d and k2 = %d.\n', k1, k2);
			fprintf('    equation1 = %d and equation2 = %d.\n', equation1, equation2);
			fprintf('    mod(equation1, 17) = %d and mod(equation2, 17) = %d.\n', eqn1mod, eqn2mod);
			results(counter, :) = [k1, k2];
			counter = counter + 1;
		end
	end
end

[Ammy]

Thank you very much for such a help.

Answer 2

[Image Analyst]

Because of missing parentheses regarding mod, and a poorly worded question, it's unclear what you want and exactly what is to be moded. But let's start here, then you can explain further.

disp('y^2 = x^3 + 14 mod 17')
y=0:16
x = (y .^ 2 - 14) .^ (1/3)
nexttile;
plot(x, y, 'b.-', 'LineWidth', 2, 'MarkerSize', 25);
grid on
xlabel('x', 'Interpreter', 'none', 'FontSize', 15);
ylabel('y', 'Interpreter', 'none', 'FontSize', 15);
for k=1:17
	left_side(k) = mod(y(k).^2,17)
	right_side(k) = mod(real(x(k).^3+14),17)
	% If I have y-coordinate from 0 to 16
	% how can I find corresponding x-coordinate that
	% satisfies 'y^2 = x^3 + 14 mod 17'
end
nexttile;
plot(right_side, left_side, 'b.-', 'LineWidth', 2, 'MarkerSize', 25);
grid on
xlabel('right_side', 'Interpreter', 'none', 'FontSize', 15);
ylabel('left_side', 'Interpreter', 'none', 'FontSize', 15);

[Ammy]

I mean in this curve 'y^2 = x^3 + 14 mod 17' we have points (1,7), (1,10), (6,3) etc if y coordinate is given how can I find x-coordinate that satifies the curve

[Image Analyst]

So if x=1 and y=7,

y^2 = 49 and mod(49,17) is 15

x^3+14 = 17 and mod(17,17) is 0

So explain again in detail why point (1,7) is a point where this condition is true.

And please make clear whether y^2 is also supposed to be moded by 17, or if it's just the right hand (x) side of the equation that's supposed to be moded with 17. In other words is it

mod(y^2, 17)

AND

mod(x.^3+14,17)

OR JUST the right hand side (because you did not use mod explicitly on the left hand side).

mod(x.^3+14,17)

In other word did you mean

y^2 = x^3 + 14 mod 17

or did you mean

y^2 mod 17 = x^3 + 14 mod 17

So, since y^2 = x.^3+14, whenever y^2 is mod 17, then x.^3+14 will also be mod 17 since y^2 = x.^3+14. They will both be mod 17 at the same time.

And finally, you might want to look at rem() rather than mod.