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Suppose that a,b,c,d are distinct integers sandwiched in an interval bounded by adjacent perfect squares, n²≤a<b<c<d≤(n+1)². Show that ad≠bc (and so the six pairwise products are distinct).
Proof:
This will be a proof by contradiction, supplied by a person on the NRICH forum who goes by the name of lebesgue.
Suppose ad=bc. Then positive coprime integers r,s exist such that a/b = c/d = r/s.
Then, for some integers k and m, a = kr, b = ks, c = mr, and d = ms.
(1) Because a<b<c<d,
m ≥ k+1, and s
(2) Since a = kr ≥ n�, it follows from the AM-GM Inequality that k+r ≥ 2n, with equality only if k=r.
Taking (1) and (2) together, we see that d = ms ≥ (k+1)(r+1) = kr+k+r+1 ≥ n�+2n+1 = (n+1)�, with equality all the way through only if m=s and k=r.
That is, d ≥ (n+1)�. Taken together with n�≤a<b<c<d≤(n+1)�, this implies that a=n� and d=(n+1)�. And this can only happen in the "equality" case, so ks=mr, which means b=c, a contradiction which completes the proof.
Nick's Mathematical Puzzles puzzle 36 is similar: If a,b,c,d are distinct positive integers with ad=bc, then a+b+c+d is composite
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