The following statement is to be proven or disproven:
Let denote the triangular numbers.
If is also a triangular number for all integer values then is a perfect square, and is a triangular number.
Can be a perfect square of an even number?
Solution
If is a triangular number, we may generalize the formula as where are natural numbers.
Substitute the above expression for to the LHS and expand the RHS. Multiply both sides by 2 to get
By equating like terms on both sides, we discover that A is a perfect square and B is a triangular number:
Since , we find that , an odd number.
Therefore, is a perfect square of an odd number, and is a triangular number.
* As an exercise, prove that .*
Check out my other notes at Proof, Disproof, and Derivation
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Comments
@Steven Zheng What about this?
Solution
If AT(n)+B is a triangular number, we may generalize the formula as AT(n)+B=2(an+p)(an+p+1) where a,n,p are natural numbers.
Substitute the above expression for T(n) to the LHS and expand the RHS. Multiply both sides by 2 to get
An2+An+2B=a2n2+a(2p+1)n+p(p+1).
By equating like terms on both sides, we discover that A is a perfect square and B is a triangular number:
A=a2=a(2p+1)
B=2p(p+1).
Since A=a2=a(2p+1), we find that a=2p+1, an odd number.
Therefore, A is a perfect square of an odd number, and B is a triangular number.
* As an extra exercise, prove that A=1+8B.*