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Revision History for A300219

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Showing entries 1-10 | older changes
Number of ways to write n^2 as x^2 + y^2 + z^2 + w^2 with x,y,z,w nonnegative integers and z <= w such that both x and 4*x - 3*y are powers of 4 (including 4^0 = 1).
(history; published version)
#20 by N. J. A. Sloane at Sat Mar 03 22:50:39 EST 2018
STATUS

proposed

approved

#19 by Zhi-Wei Sun at Sat Mar 03 22:35:11 EST 2018
STATUS

editing

proposed

#18 by Zhi-Wei Sun at Sat Mar 03 22:32:41 EST 2018
COMMENTS

In arXiv:1701.05868 the author proved that any positive integer square can be written as (4^k)^2 + x^2 + y^2 + z^2 with k,x,y,z nonnegative integers, and for each square greater than one r = 0,1 and n > r we can be written write n^2 as (2*4^k(2k+r))^2 + x^2 + y^2 + z^2 with k,x,y,z nonnegative integers.

Discussion
Sat Mar 03
22:35
Zhi-Wei Sun: Make the comments more compact.
#17 by Zhi-Wei Sun at Sat Mar 03 22:28:55 EST 2018
COMMENTS

(ii) Let r be 0 or 1, and let n > r. Then n^2 can be written as x^2 + y^2 + z^2 + w^2 with x,y,z,w nonnegative integers such that both x and x + 3*y belong to the set {2^(2k+r): k = 0,1,2,...}, unless n has the form 2^(2k+r)*81503 with k a nonnegative integer and hence n^2 = (2^(2k+r)*28^2)^2 + (2^(2k+r)*80)^2 + (2^(2k+r)*55937)^2 + (2^(2k+r)*59272)^2 with 2^(2k+r)*28^2 = 2^r*(2^k*28)^2 and 2^(2k+r)*28^2 + 3*(2^(2k+r)*80) = 2^(2(k+5)+r). So n^2 we always can be written write n^2 as x^2 + y^2 + z^2 + w^2 with x,y,z,w nonnegative integers such that x/2^r is a square and (x + 3*y )/2^r is a power of 4.

#16 by Zhi-Wei Sun at Sat Mar 03 22:24:36 EST 2018
COMMENTS

(ii) A positive integer square Let r be 0 or 1, and let n > r. Then n^2 can be written as x^2 + y^2 + z^2 + w^2 with x,y,z,w nonnegative integers such that both x and x + 3*y are powers of 4 (including 4belong to the set {2^(2k+r): k = 0 = ,1), ,2,...}, unless n has the form 42^k(2k+r)*81503 with k a nonnegative integer and hence n^2 = (42^k(2k+r)*28^2)^2 + (42^k(2k+r)*80)^2 + (42^k(2k+r)*55937)^2 + (42^k(2k+r)*59272)^2 with 42^k(2k+r)*28^2 = 2^r*(2^k*28)^2 and 42^k(2k+r)*28^2 + 3*(42^k(2k+r)*80) = 42^(2(k+5)+r). Also, a square So n^2 > 1 can be written as x^2 + y^2 + z^2 + w^2 with x,y,z,w nonnegative integers such that both x /2^r is a square and x + 3*y belong to the set {2^(2k+1): k = 0,1,...}, unless n has the form 2^(2k+1)*81503 with k is a nonnegative integer and hence n^2 = (2^(2k+1)*28^2)^2 + (2^(2k+1)*80)^2 + (2^(2k+1)*55937)^2 + (2^(2k+1)*59272)^2 with 2^(2k+1)*28^2 = 2*(2^k*28)^2 and 2^(2k+1)*28^2 + 3*(2^(2k+1)*80) = 2^(2(k+5)+1)power of 4.

STATUS

approved

editing

#15 by Bruno Berselli at Thu Mar 01 03:42:43 EST 2018
STATUS

proposed

approved

#14 by Zhi-Wei Sun at Thu Mar 01 03:15:18 EST 2018
STATUS

editing

proposed

#13 by Zhi-Wei Sun at Thu Mar 01 03:15:13 EST 2018
COMMENTS

Conjecture: (i) a(n) > 0 for all n > 0, and a(n) = 1 only for n = 4^k*m (k = 0,1,2,... and m = 1, 2, 3, 5, 15, 37, 83, 263). Also, for each n = 2,3,... we can write n^2 as x^2 + y^2 + z^2 + w^2 with x,y,z,w nonnegative integers such that both x and 4*x - 3*y lie in the set {2^(2k+1): k = 0,1,2,...}.

(ii) A positive integer square n^2 can be written as x^2 + y^2 + z^2 + w^2 with x,y,z,w nonnegative integers such that both x and x + 3*y are powers of 4 (including 4^0 = 1), unless n has the form 4^k*81503 with k a nonnegative integer and hence n^2 = (4^k*28^2)^2 + (4^k*80)^2 + (4^k*55937)^2 + (4^k*59272)^2 with 4^k*28^2 = (2^k*28)^2 and 4^k*28^2 + 3*(4^k*80) = 4^(k+5). Also, a square n^2 > 1 can be written as x^2 + y^2 + z^2 + w^2 with x,y,z,w nonnegative integers such that both x and x + 3*y belong to the set {2^(2k+1): k = 0,1,...}, unless n has the form 2^(2k+1)*81503 with k a nonnegative integer and hence n^2 = (2^(2k+1)*28^2)^2 + (2^(2k+1)*80)^2 + (2^(2k+1)*55937)^2 + (2^(2k+1)*59272)^2 with 2^(2k+1)*28^2 = 2*(2^k*28)^2 and 2^(2k+1)*28^2 + 3*(2^(2k+1)*80) = 2^(2(k+5)+1).

#12 by Zhi-Wei Sun at Thu Mar 01 03:04:36 EST 2018
COMMENTS

Conjecture: (i) a(n) > 0 for all n > 0, and a(n) = 1 only for n = 4^k*m (k = 0,1,2,... and m = 1, 2, 3, 5, 15, 37, 83, 263). Also, for each n = 2,3,... we can write n^2 as x^2 + y^2 + z^2 + w^2 with x,y,z,w nonnegative integers such that both x and 4*x - 3*y belong to lie in the set {2^(2k+1): k = 0,1,2,...}.

In arXiv:1701.05868 the author proved that any positive integer square can be written as (4^k)^2 + x^2 + y^2 + z^2 with k,x,y,z nonnegative integers, and each square greater than one can be written as (2*4^k)^2 + x^2 + y^2 + z^2 with k,x,y,z nonnegative integers.

#11 by Zhi-Wei Sun at Thu Mar 01 03:02:09 EST 2018
COMMENTS

Conjecture: (i) a(n) > 0 for all n > 0, and a(n) = 1 only for n = 4^k*m (k = 0,1,2,... and m = 1, 2, 3, 5, 15, 37, 83, 263). Also, for each n = 2,3,... we can write n^2 as x^2 + y^2 + z^2 + w^2 with x,y,z,w nonnegative integers such that both x and 4*x - 3*y belong to the set {2^(2k+1): k = 0,1,2,...}.