1. ## Euler's Criterion

I am trying to show that x^2 == 25(mod 997) has a solution, but I can't seem to get there.

2. x=5?

3. Yeah, thanks. Caught onto that after I belatedly realized that 997 is prime. Thank you!

4. Originally Posted by tarheelborn
Yeah, thanks. Caught onto that after I belatedly realized that 997 is prime. Thank you!
It doesn't matter in this case that 997 is prime; you'd still have the solution $\displaystyle x=\pm 5$ (although perhaps you'd have others as well).

5. But since 997 is prime, wouldn't Lagrange's theorem apply to show that there are ONLY 2 solutions?

6. Originally Posted by tarheelborn
But since 997 is prime, wouldn't Lagrange's theorem apply to show that there are ONLY 2 solutions?

What has Lagrange to do here? Since 997 is a prime there are only two solutions because any polynomial p(x) ( e.g., $\displaystyle x^2-25=0$ ) over a field can have at most $\displaystyle \deg(p(x))$ different roots in some extension of the field.

Tonio

7. Originally Posted by Bruno J.
It doesn't matter in this case that 997 is prime; you'd still have the solution $\displaystyle x=\pm 5$ (although perhaps you'd have others as well).
Sure, I see that; I was just trying to clarify your statement that perhaps you would have other solutions as well. Thank you so much for your help!