# Proving a multi-variable function is injective

I am required to prove this statement if it is true, or disprove it if it isn't. The function that I am working with is as follows: $$p: \mathbb{N} × \mathbb{N} \to \mathbb{N}, p(a, b) = \frac{ab(b+1)}{2}$$.

I am suspicious that this function is not injective, but I can't seem to find a counter-example, so I tried negating the definition of an injective function, that is $$\forall x_1,y_1, x_2, y_2 \in \mathbb{N} × \mathbb{N}, p(x_1, y_1) = p(x_2, y_2) \Rightarrow (x_1, y_1) \ne (x_2, y_2)$$ to prove the negation is true, which would imply the original statement is false. However, I am not sure if that is right.

What I did so far is, I simply supposed that $$p(x_1, y_1) = p(x_2, y_2)$$, so I got $$\frac{x_1y_1(y_1+1)}{2} = \frac{x_2y_2(y_2+1)}{2}$$ $$\Leftrightarrow x_1y_1(y_1+1) = x_2y_2(y_2+1)$$. But once I got here, I don't really know what to do next.

$$p(a_1,3)=6a_1$$, and $$p(a_2,4)=10a_2$$. To prove the function is not injective it suffices to find $$a_1, a_2 \in \mathbb{N}$$ such that $$6a_1=10a_2$$.
Hint: Consider the LCM of $$6$$ and $$10$$.
• Would $a_1 = \frac{30}{6} = 5, a_2 = \frac{30}{10} = 3$
• Yes with this choice of $a_1$ and $a_2$, one sees $p(5,3)=p(3,4)$, but $(5,3)\neq(3,4)$. Mar 11, 2020 at 21:11