# 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
5 years, 3 months ago

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- 4 years, 10 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 $A = 1+ 8B$.*

- 5 years, 2 months ago