Several different positive integers lie strictly between two successive squares. Prove that their pairwise products are also different.

Let the numbers be $n$ and $n+1$. So, their squares are $n^2$ and $n^2 + 2n+1$.

I haven't found an idea on how to solve this. Could you suggest some ideas or give hints on how to solve this?



Partial solution sketch

Take any $m,n,a,b,c,d$ such that $m^2 < a < b \le c < d < n^2$. If $ad = bc$ then let $p,q,r,s$ be such that $(a,b,c,d) = (pq,pr,qs,rs)$ (say by letting $p = gcd(a,b)$ and $q = \frac{a}{p}$ and $r = \frac{b}{p}$; and you can work out $s$ and prove that $s$ is an integer). Then note that $q < r$ and $ad = bc = pqrs$ and we have the following cases:

  1. If $p \le q < r \le s$ then $a = pq \le q^2 < r^2 \le rs = d$.

  2. If $q \le p \le r$ then $a = pq \le p^2 \le pr = b$.

  3. If $q \le s \le r$ then $c = qs \le s^2 \le rs = d$.

  4. If $q < r < p < s$ then ??? (And symmetrically if $p < s < q < r$.)

  • $\begingroup$ I took the pairs $n^2 + k_1, n^2 + k_2$ and $ n^2 + k_3, n^2 + k_4$ as the numbers and have reached $$n^2(k_1+k_2-k_3-k_4) = k_3k_4-k_1k_2$$ Now how do I find a perfect square? $\endgroup$ – TheRandomGuy Apr 4 '16 at 10:52
  • $\begingroup$ Could you elaborate on the hint please? $\endgroup$ – shardulc Apr 4 '16 at 11:32
  • $\begingroup$ @shardulc,Dhruv: Sorry I missed a case and couldn't finish. Here's what I have. $\endgroup$ – user21820 Apr 4 '16 at 11:52

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