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I need to prove that $f: \mathbb{Z}^+ \times \mathbb{Z}^+ \rightarrow \mathbb{Z}^+$, $f(x) = \frac{(m+n-2)(m+n-1)}{2} + m$ is injective.

Start off by $(a,b) \ne (c,d) \implies f(a,b) \ne f(c,d)$

After plugging this in, I get: $$\frac{(a+b-2)(a+b-1)}{2}+a = \frac{(c+d-2)(c+d-1)}{2}+c$$

$$a^2 + 2ab - a - 3b + b^2 = c^2 + 2cd - c - 3d + d^2$$

Can anyone give a hint on how to proceed from here?

john doe
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  • Hint: Can you show that if you're given the value of $f(x)$, then $m+n$ is uniquely determined? – Calvin Lin Jun 14 '23 at 14:13
  • @CalvinLin I'm afraid not. Would appreciate an explanation because it seems interesting – john doe Jun 14 '23 at 14:17
  • Injectivity is implication $x\ne y \Rightarrow f(x) \ne f(y)$ or, it's same, $f(x) =f(y) \Rightarrow x=y$ https://en.wikipedia.org/wiki/Injective_function. What you have "$f(a,b) \ne f(c,d) \implies (a,b) \ne (c,d)$" is only definition of function. – zkutch Jun 14 '23 at 14:21
  • @zkutch The first definition of injectivity is what I was operating on, but I made an error in typing. Fixed. – john doe Jun 14 '23 at 14:30
  • @CalvinLin It does help, and using the same method I arrived at $m-p = ad + \frac{d(d-3)}{2}$.

    However, the original author then stated that this should be $\ge a+1$ and I'm not sure how they arrived to that conclusion. What would be the analogous inequality in my case, and why?

     – john doe Jun 14 '23 at 15:31

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