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

## Comments

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TopNewest@Steven Zheng What about this?

– Megh Choksi · 2 years, 2 months agoLog in to reply

SolutionIf \(AT(n)+B\) 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 \(A = 1+ 8B\).*– Steven Zheng · 2 years, 7 months agoLog in to reply