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

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?

Thanks.

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 $p < s < q < r$.)

• 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? – TheRandomGuy Apr 4 '16 at 10:52
• Could you elaborate on the hint please? – shardulc Apr 4 '16 at 11:32
• @shardulc,Dhruv: Sorry I missed a case and couldn't finish. Here's what I have. – user21820 Apr 4 '16 at 11:52