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Question
if n>4 is a composite number, show that n|(n-1)! Conclude that (n-1)! is not congruent to -1(mod n).
Hello Surya
Unless n is a square, it can be factorised, and both of these factors are less than n, and can be found as factors of (n-1)!.
So what happens when n is a square. n = 4 is an exception. But for any other square,
k^2, both k and 2k are factors of (k^2-1)!
Best wishes
vijilant
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Vijilant
Expertise
Most questions on number theory, divisibility, primes, Euclidean algorithm, Fermat`s theorem, Wilson`s theorem, factorisation, euclidean algorithm, diophantine equations, Chinese remainder theorem, group theory, congruences, continued fractions.
Experience
Teacher of math for 50 years
Organizations
ATL
Publications
Journal of mathematics and its applications
Education/Credentials
BSc Hons Liverpool
Awards and Honors
State Scholarship 1955
Past/Present Clients
I taught John Birt, former Director of the BBC in 1961. His homework book was the most perfect I have ever marked. And also the most neat. I could tell he was destined for great things.
One of my classmates was the poet Roger McGough, and I have a mention in his autobiography.
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