Triangular Numbers

The following statement is to be proven or disproven:

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

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

Can AA be a perfect square of an even number?

Solution

If AT(n)+BAT(n)+B is a triangular number, we may generalize the formula as AT(n)+B=(an+p)(an+p+1)2AT(n)+B = \frac{(an+p)(an+p+1)}{2} where a,n,pa,n,p are natural numbers.

Substitute the above expression for T(n)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).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=a2=a(2p+1)A = {a}^{2} = a(2p+1)

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

Since A=a2=a(2p+1)A = {a}^{2} = a(2p+1), we find that a=2p+1a = 2p+1, an odd number.

Therefore, AA is a perfect square of an odd number, and BB is a triangular number.

* As an exercise, prove that A=1+8BA = 1+ 8B.*

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

#NumberTheory #TriangularNumbers

Note by Steven Zheng
6 years, 11 months ago

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Comments

@Steven Zheng What about this?

U Z - 6 years, 5 months ago

Solution

If AT(n)+BAT(n)+B is a triangular number, we may generalize the formula as AT(n)+B=(an+p)(an+p+1)2AT(n)+B = \frac{(an+p)(an+p+1)}{2} where a,n,pa,n,p are natural numbers.

Substitute the above expression for T(n)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).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=a2=a(2p+1)A = {a}^{2} = a(2p+1)

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

Since A=a2=a(2p+1)A = {a}^{2} = a(2p+1), we find that a=2p+1a = 2p+1, an odd number.

Therefore, AA is a perfect square of an odd number, and BB is a triangular number.

* As an extra exercise, prove that A=1+8BA = 1+ 8B.*

Steven Zheng - 6 years, 10 months ago
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