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How come, for any n > 2, the nth triangular number + the nth square

by goldust
Tags: number, square, triangular
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goldust
#1
Nov5-13, 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.
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Mentallic
#2
Nov5-13, 09:54 AM
HW Helper
P: 3,531
The nth triangle number Tn is

[tex]T_n = \frac{n(n+1)}{2}[/tex]

and the nth square number Sn 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?
goldust
#3
Nov5-13, 11:09 AM
P: 85
Can you now show why this expression must be composite (not prime) for all n?
Well, it wouldn't be true for all n. Only for n > 2, assuming my statement is correct.

goldust
#4
Nov5-13, 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.
Mentallic
#5
Nov5-13, 08:40 PM
HW Helper
P: 3,531
Quote Quote by goldust View Post
Well, it wouldn't be true for all n. Only for n > 2, assuming my statement is correct.
Well, ignoring the fact that you included the criteria that n>2 in your proof below, [itex]n\leq 2[/itex] would also spit out composite numbers, right? Your proof only considers that n is even or odd which means that all integers [itex]n\leq 2[/itex] would also be involved. The only reason to restrict yourself to n>2 is such that we have a meaningful square and triangle number.

Quote Quote by goldust View Post
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.
Adding to the end of that: because we then have a product of two integers, mainly n and [itex]\frac{3n+1}{2}[/itex].

Quote Quote by goldust View Post
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.
This is incorrect. If [itex]n(3n+1)[/itex] is even, then [itex]\frac{n}{2}(3n+1)[/itex] isn't necessarily even, but rather an integer. But most importantly, you haven't proven that the expression is a product of two integers and hence composite.


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.
goldust
#6
Nov5-13, 09:28 PM
P: 85
Many thanks.
willem2
#7
Nov6-13, 01:55 AM
P: 1,396
Quote Quote by Mentallic View Post
Well, ignoring the fact that you included the criteria that n>2 in your proof below, [itex]n\leq 2[/itex] would also spit out composite numbers, right? Your proof only considers that n is even or odd which means that all integers [itex]n\leq 2[/itex] would also be involved. The only reason to restrict yourself to n>2 is such that we have a meaningful square and triangle number.
You do get a prime number for n=1 or n=2, and the proof also uses the fact that n>2, when it says
When n is odd and more than 2, 1 / 2 * (3n + 1) is a whole number bigger than 1,
so I don't see what the problem is here.
Mentallic
#8
Nov6-13, 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.
goldust
#9
Nov6-13, 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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