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Given, $$X \bmod 7 = 3, Y \bmod 7 = 2, Y^2 = X^3 + 9$$

How to evaluate X and Y? Is there any rule solving this type of problem for any given modulus value and equation?

Maximilien Mackie
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Asif Iqbal
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    If $X \bmod 7 = 3$ then $X^3 \bmod 7 = 3^3 \bmod 7=27 \bmod 7=6$ , similarly, if $Y \bmod 7 = 2$, $ Y^2 \bmod 7 = 2^2 = 4$. But then $X^3 +9 \bmod 7 = 6+9 \bmod 7= 15 \bmod 7 = 1 \neq 4 = Y^2 \bmod 7$ so I don't think there are any solution – Anthony Sep 26 '24 at 09:41
  • By Congruence Laws $\bmod 7!:\ \color{#c00}{x\equiv 3},\ \color{#0a0}{y\equiv 2}\Rightarrow \color{#0a0}{y}^2\equiv \color{#0a0}2^2\equiv 4,,$ $\color{#c00}x^3\equiv \color{#c00}3^3\equiv -1,,$ but $,y^2\equiv 4,$ is not congruenct to $,x^3+9\equiv -1+9\equiv 1\ \ $ – Bill Dubuque Sep 26 '24 at 16:33

1 Answers1

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Given the values you have specified: $$Y^2=4$$ and $$X^3+9=1$$

You are therefore asking for a solution of the equation $$4=1$$

Martin Kochanski
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