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# Triangular Numbers

The following statement is to be proven or disproven:

Let $T(n)=\frac{n(n+1)}{2}$ denote the triangular numbers.

If $$AT(n)+B$$ is also a triangular number for all integer values $$(A,B)$$ then $$A$$ is a perfect square, and $$B$$ is a triangular number.

Can $$A$$ be a perfect square of an even number?

Solution

If $$AT(n)+B$$ is a triangular number, we may generalize the formula as $AT(n)+B = \frac{(an+p)(an+p+1)}{2}$ 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

$A{n}^{2}+An+2B = {a}^{2}{n}^{2}+a(2p+1)n+p(p+1).$

By equating like terms on both sides, we discover that A is a perfect square and B is a triangular number:

$A = {a}^{2} = a(2p+1)$

$B = \frac{p(p+1)}{2}.$

Since $$A = {a}^{2} = a(2p+1)$$, 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 exercise, prove that $$A = 1+ 8B$$.*

Check out my other notes at Proof, Disproof, and Derivation

Note by Steven Zheng
3 years, 3 months ago

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- 2 years, 9 months ago

Solution

If $$AT(n)+B$$ is a triangular number, we may generalize the formula as $AT(n)+B = \frac{(an+p)(an+p+1)}{2}$ 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

$A{n}^{2}+An+2B = {a}^{2}{n}^{2}+a(2p+1)n+p(p+1).$

By equating like terms on both sides, we discover that A is a perfect square and B is a triangular number:

$A = {a}^{2} = a(2p+1)$

$B = \frac{p(p+1)}{2}.$

Since $$A = {a}^{2} = a(2p+1)$$, 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 $$A = 1+ 8B$$.*

- 3 years, 2 months ago