Proposition needed to be proven

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The discussion revolves around a mathematical proof concerning the proposition that for any positive integer n, there are no positive integers a, b, c such that 4n + 3 equals 5^a * 13^b * 17^c. Participants clarify the modular arithmetic involved, noting that while the left-hand side is 3 mod 4, the right-hand side equals 1 mod 4, leading to a contradiction. They also discuss the importance of testing small values of n to find primes generated by the sequence defined by 4n + 3. The conversation touches on the notation \mathbb{Z^{*}}, which denotes non-negative integers. Ultimately, the proof is confirmed to be complete through the modular analysis presented.
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A relatively lengthy proof I am writing for an assignment leads me to a proposition (which I need to turn into a lemma for the proof to be complete) conjecturing that for any n \in \mathbb{Z^{*}}, there are no a, b, c \in \mathbb{Z^{*}} such that 4n + 3 = 5^{a}13^{b}17^{c}. I haven't been able to find to tackle the problem, any suggestions?
 
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I don't understand you question...do you mean that the sequence defined by a_n = 4n +3 , n \in \mathbb{Z} must generate some primes for n other than 5, 13, 17 and 21?

The digits 4 and 3 add up to 7, a prime. So a value of n that would keep the digits the same already rules the number out for a heap of divisibility tests for small numbers. Since it only rules out small numbers, try small values of n. n=1, a_1 = 7, prime. n=10, a_10 = 43, prime.

I don't think that's what you are asking, because I'm sure you would have spotted n=1 straight away.
 
What do you mean by "4n + 3 must have primes other than 5, 13, 17 and 21"?

21 isn't prime :confused:

5, 13, 17, 21 are all of the form 4n+1 not 4n+3 :confused:
 
Sorry for stating that 21 is prime, it was 4 am here, easy to say gibberish at this time. That said, I have rectified the original question, so please read it.
 
(4n+1)(4k+1) = 16nk+4(n+k)+1 = 4m+1. The form is preserved under multiplication.
 
OK, I read it, but robert Ihnot got there first...

5 = 13 = 17 = 1 mod 4, so 5^a 13^b 17^c = 1 mod 4.
 
Thanks allot, the proof is complete. :smile:
 
work mod 4. the lhs is 3 mod 4. what aboutn the rhs? 5=1 mod 4, 13=1 mod 4, and also 17=1 mod 4, so the lhs =3 and the rhs =1.
 
Sorry to sound ignorant but what does the "*" represent in \mathbb{Z}^{*}
 
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  • #10
uart said:
Sorry to sound ignorant but what does the "*" represent in \mathbb{Z}^{*}

Non-negative integers.
 
  • #11
"Non-negative". Ok thanks.
 
  • #12
I'm curious as to why you ask this question, Werg22. I vaguely remember using/reading this lemma, as proven by robert Ihnot here, in a maths book. It was titled "Proofs from the BOOK" or something along those lines. I believe the chapter I found it in was something on the representation of integers as sum of primes or something like that.
 

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