proofs

Binomial coefficients are (almost) never powers 15
So for each i the number of multiples of p i among n, . . . , n−k+1, and
hence among the a j ’s, is bounded by ⌊ k p i ⌋ + 1. This implies that the exponent
of p in a 0 a 1 · · · a k−1 is at most
∑l−1
(⌊ k
⌋ )
p i + 1
i=1
with the reasoning that we used for Legendre’s theorem in Chapter 2. The
only difference is that this time the sum stops at i = l − 1, since the a j ’s
contain no l-th powers.
Taking both counts together, we find that the exponent of p in v l is at most
∑l−1
(⌊ k
⌋ )
p i + 1 − ∑ ⌊ k
⌋
p i i=1
i≥1
≤ l − 1,
and we have our desired contradiction, since v l is an l-th power.
This suffices already to settle the case l = 2. Indeed, since k ≥ 4 one of
the a i ’s must be equal to 4, but the a i ’s contain no squares. So let us now
assume that l ≥ 3.
(4) Since k ≥ 4, we must have a i1 = 1, a i2 = 2, a i3 = 4 for some i 1 , i 2 , i 3 ,
that is,
n − i 1 = m l 1 , n − i 2 = 2m l 2 , n − i 3 = 4m l 3 .
We see that our analysis so far agrees
with `50
3
´
= 140 2 , as
50 = 2 · 5 2
49 = 1 · 7 2
48 = 3 · 4 2
and 5 · 7 · 4 = 140.
We claim that (n − i 2 ) 2 ≠ (n − i 1 )(n − i 3 ). If not, put b = n − i 2 and
n − i 1 = b − x, n − i 3 = b + y, where 0 < |x|, |y| < k. Hence
b 2 = (b − x)(b + y) or (y − x)b = xy,
where x = y is plainly impossible. Now we have by part (1)
|xy| = b|y − x| ≥ b > n − k > (k − 1) 2 ≥ |xy|,
which is absurd.
So we have m 2 2 ≠ m 1 m 3 , where we assume m 2 2 > m 1 m 3 (the other case
being analogous), and proceed to our last chains of inequalities. We obtain
2(k − 1)n > n 2 − (n − k + 1) 2 > (n − i 2 ) 2 − (n − i 1 )(n − i 3 )
= 4[m 2l
2 − (m 1 m 3 ) l ] ≥ 4[(m 1 m 3 + 1) l − (m 1 m 3 ) l ]
≥
4lm l−1
1 m l−1
3 .
Since l ≥ 3 and n > k l ≥ k 3 > 6k, this yields
2(k − 1)nm 1 m 3 > 4lm l 1 ml 3 = l(n − i 1 )(n − i 3 )
> l(n − k + 1) 2 > 3(n − n 6 )2 > 2n 2 .