Download Solution #13

Problem of the Week #13
2/20/2017 to 3/5/2017
Let x be an integer. Show that
(x + 1)2 + (x + 2)2 + · · · + (x + 99)2 6= ab
for any integers a, b with b ≥ 2.
Solution: Set F (x) := (x + 1)2 + (x + 2)2 + · · · + (x + 99)2 . Then
F (x) = 99x2 + 2(1 + 2 + · · · + 99)x + (12 + 22 + · · · + 992 )
100 · 99
(2 · 100 − 1) · 100 · 99
2
= 99x + 2
+
2
6
2
= 33(3x + 300x + 9950).
We know that 3 divides 33, but 32 does not. Further, 3 does not divide 3x2 + 300x + 9950,
since 3 divides 3x2 and 300x but not 9950. Thus, F (x) is divisible by 3, but it is not divisible
by 32 .
Now, if such an a and b did exist, then any prime which divides ab appears at least b ≥ 2
times in the prime factorization of ab . Thus, since 3 is a prime factor of F (x) which appears
only once in the prime factorization of F (x), no such a and b exist.
Solutions for this problem were submitted by M.V. Channakeshava (Bengaluru,
India), Rob Hill (Gambrills, Maryland), Lincoln James (Chicago, IL), Kipp
Johnson (Beaverton, OR), Hari Kishan (India), Benjamin Phillabaum (Northbrook, IL), Victoria Rose (Northbrook, IL), and Luciano Santos (Lisboa, Portugal).