804 Sums of consecutive squares
k =6, 8 , 10 , 13 , 17 , 18 , 20 , 21 , 22 , 26 , 27 , 30 , 32 ,
34 , 35 , 37 , 40 , 41 , 42 , 44 , 45 , 46 , 48 ,...5.Supposep=2k+1is a prime. If the Legendre symbol(
−^13 k(k+1)
p)
=
− 1 , then the equation(Ek)hasnosolution. Verify that for the fol-
lowing values ofk< 50 , the equation(Ek)has no solution:1 , 2 , 3 , 8 , 9 , 14 , 15 , 20 , 21 , 26 , 33 , 39 , 44.6.Fork≤ 50 , it remains to consider(Ek)for the following values of
k:
5 , 7 , 11 , 16 , 19 , 23 , 25 , 28 , 29 , 31 , 36 , 38 , 43 , 47 , 49.
Among these, only fork=5, 11 , 16 , 23 , 29 are the equations(Ek)
solvable.7.Work out 5 sequences of 23 consecutive integers whose squares
add up to a square in each case.
Answer:72 +8^2 +···+29^2 =92^2 ;
8812 + 882^2 +···+ 903^2 = 4278^2 ;
427872 + 42788^2 +···+ 42809^2 = 205252^2 ;
20534012 + 2053402^2 +···+ 2053423^2 = 9847818^2 ;
·········8.Consider the equation(E 36 ):u^2 − 73 v^2 =12· 37 · 73. This equation
does in fact have solutions(u, v) = (4088,478),(23360,2734).
The fundamental solution of the Pell equationx^2 − 73 y^2 =1being
(a, b)= (2281249, 267000), we obtain two sequences of solutions
of(E 73 ):
Answer:
(4088,478),(18642443912,2181933022),(85056113063608088,9955065049008478),...
(23360,2734),(106578370640,12474054766),(486263602888235360,56912849921762734),...This means, for example, the sum of the squares of the 73 numbers
with center 478 (respectively 2734) is equal to the square of 4088
(respectively 23360).