29.2 Sums of consecutive squares: odd number case 803
29.2 Sums of consecutive squares: odd number case ....
Suppose the sum of the squares of 2 k+1consecutivepositiveintegers
is a square. If the integers areb, b± 1 ,...,b±k. We require
(2k+1)b^2 +
1
3
k(k+ 1)(2k+1)=a^2
for an integera. From this we obtain the equation
a^2 −(2k+1)b^2 =
1
3
k(k+ 1)(2k+1). (Ek)
1.Suppose 2 k+1is a square. Show that(Ek)has solution only when
k=6m(m+)for some integersm> 1 , and=± 1. In each
case, the number of solutions isfinite.
Number of solutions of(Ek)when 2 k+1is a square
2 k+ 1 25 49 121 169 289 361 529 625 841 961 ...
01127353310 ...
2.Find theuniquesequence of 49 (respectively 121) consecutive pos-
itive integers whose squares sum to a square.
3.Find the two sequences of 169 consecutive squares whose sums are
squares.
4.Suppose 2 k+1is not a square. Ifk+1is divisible9=3^2 or
by any prime of the form 4 k+3≥ 7 , then the equation(Ek)has
no solution. Verify that for the following values ofk< 50 , the
equation(Ek)has no solution: