29.3 Sums of consecutive squares: even number case 805
29.3 Sums of consecutive squares: even number case ....
Suppose the sum of the squares of the 2 kconsecutive numbers
b−k+1,b−k+2,...,b,...,b+k− 1 ,b+k,is equal toa^2. This means
(2a)^2 − 2 k(2b+1)^2 =2 k
3(4k^2 −1). (Ek′)Note that the numbers 2 k, 4 k^2 − 1 are relatively prime.
1.Show that the equation(E′k)has no solution if 2 kis a square.2.Suppose 2 kis not a square. Show that if 2 k+1is divisible by 9,
or by any prime of the form 4 k+1, then the equation(E′k)has no
solution.3.Fork ≤ 50 , the equation(Ek′)has no solution for the following
values ofk:k =3, 4 , 5 , 9 , 11 , 13 , 15 , 17 , 21 , 23 , 24 , 27 , 29 , 31 , 33 ,
35 , 38 , 39 , 40 , 41 , 45 , 47 , 49.4.Letkbe a prime. The equation(Ek′)can be written as(2b+1)^2 − 2 ky^2 =−4 k^2 − 1
3.
By considering Legendre symbols, the equation(Ek′)has no solu-
tion for the following values ofk≤ 50 :k=5, 7 , 17 , 19 , 29 , 31 , 41 , 43.5.Excluding square values of 2 k< 100 , the equation(Ek′)has solu-
tions only fork=1, 12 , 37 , 44.6.Show that (34, 0), (38, 3), (50, 7) are solutions of(E” 12 ). Construct
from them three infinite sequences of expressions of the sum of 24
consecutive squares as a square.