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

[goldust]

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.

 

[Mentallic]

The n^th triangle number T_n is

T_n = \frac{n(n+1)}{2}

and the n^th square number S_n is

S_n = n^2

And so
T_n+S_n = \frac{n(n+1)}{2}+n^2=\frac{n(3n+1)}{2}

Can you now show why this expression must be composite (not prime) for all n?

 

[goldust]

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]

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]

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, n\leq 2 would also spit out composite numbers, right? Your proof only considers that n is even or odd which means that all integers n\leq 2 would also be involved. The only reason to restrict yourself to n>2 is such that we have a meaningful square and triangle number.

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 \frac{3n+1}{2}.

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 n(3n+1) is even, then \frac{n}{2}(3n+1) 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]

Many thanks.

 

[willem2]

Well, ignoring the fact that you included the criteria that n>2 in your proof below, n\leq 2 would also spit out composite numbers, right? Your proof only considers that n is even or odd which means that all integers n\leq 2 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]

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]

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!