000RM.dvi

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: