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Suppose that x and y are integers with x = 3 mod 11 and y = 6 mod 11.

Solve for z: y^2 = 5x + z mod 11.

What I have done: By just doing the algebra I get z = 21 and then I do 21 mod 11 and get 10. Is that correct? I feel like that is too simple.

Note: All of the = should be the congruent sign.

1 Answers1

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Yes, it is correct: $\ {\rm mod}\ 11\!:\ x\equiv 3,\ y\equiv 6\,\Rightarrow\, z \equiv y^2-5x\equiv 6^2-5\cdot 3\equiv 36-15\equiv 21$

The argument employs the Congruence Sum and Product Rules. Be sure you understand them.

Bill Dubuque
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