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How come, for any n > 2, the nth triangular number + the nth square 
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#1
Nov513, 08:57 AM

P: 85

number cannot be prime? I have checked this for n from 3 to 53,509, the latter being the limit for unsigned int. I believe this is true, and I thereby claim that this is a true statement. However, I don't see any obvious explanation for it.



#2
Nov513, 09:54 AM

HW Helper
P: 3,531

The n^{th} triangle number T_{n} is
[tex]T_n = \frac{n(n+1)}{2}[/tex] and the n^{th} square number S_{n} is [tex]S_n = n^2[/tex] And so [tex]T_n+S_n = \frac{n(n+1)}{2}+n^2=\frac{n(3n+1)}{2}[/tex] Can you now show why this expression must be composite (not prime) for all n? 


#3
Nov513, 11:09 AM

P: 85




#4
Nov513, 12:38 PM

P: 85

How come, for any n > 2, the nth triangular number + the nth square
When n is odd and more than 2, 1 / 2 * (3n + 1) is a whole number bigger than 1, and so the result is composite. When n is even and more than 2, n * (3n + 1) is even, so (n / 2) * (3n + 1) is even. Therefore the sum is composite for all n > 2 is correct. Many thanks for the help.



#5
Nov513, 08:40 PM

HW Helper
P: 3,531

Also, while it's not absolutely necessary, when you consider n to be even, you could let n=2k for some integer k and substitute that into your expression, then show that the result is composite, and similarly for n odd, let n=2k+1. 


#6
Nov513, 09:28 PM

P: 85

Many thanks.



#7
Nov613, 01:55 AM

P: 1,396




#8
Nov613, 05:03 AM

HW Helper
P: 3,531

Haha yeah I thought about it while out today and realized the criteria n>2 is necessary, which goldust even incorporated into his proof! Sorry about that goldust.



#9
Nov613, 08:59 AM

P: 85

Many thanks for the help! The proof is a bit trickier than I initially thought. When n is even and more than 2, n / 2 is an integer more than 1, and 3n + 1 is also an integer more than 1, so n / 2 * (3n + 1) ends up being divisible by both n / 2 and 3n + 1. Cheers!



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