[number theory] x²-a = 0 no solution => n not prime

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SUMMARY

The discussion centers on proving that the number n = 3^{100} + 2 is not prime by analyzing the equation x² - 53 ≡ 0 mod n, which has no solution. The key argument involves the Jacobi symbol, where if n were prime, it would imply that (53/n) = -1, leading to a contradiction with the expression 53^{(n-1)/2} mod n. The conversation suggests using the Lucas primality test as a potential method for further analysis.

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Homework Statement


Define [itex]n = 3^{100}+2[/itex]. Suppose [itex]x^2-53 \equiv 0 \mod n[/itex] has no solution. Prove that n is not prime.

Homework Equations


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The Attempt at a Solution


Well, I suppose that I'll have to prove that some identity which should be true for n prime is not satisfied in the above case. The only relevant thing that I can think of is that if n were prime, then [itex]\left( \frac{53}{n} \right) \equiv 53^{ \frac{n-1}{2} } \mod n[/itex] (the first symbol denoting the Jacobi symbol). From now on assume n is prime; I try to find a contradiction.

The fact that the stated equation has no solution, is translated into [itex]\left( \frac{53}{n} \right) = -1[/itex]. So assuming n is prime, we have that [itex]-1 \equiv 53^{ \frac{n-1}{2} } \mod n[/itex]. However, I don't see how to arrive at a contradiction, nor do I see another way to approach the problem...
 
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I don't know if it's a problem you're already done with, but try applying the lucas primality test?
 

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