• #### Sums of 2 squares ( == multiple of 6 )

From henhanna@gmail.com@21:1/5 to All on Tue Jul 5 08:28:57 2022
Of these numbers ( 2 sets )

Set1= (6, 12, 18, 24, . . . , 150, 156) = (all multiples of 6 under 160)

Set2= (1200, 1206, 1212, 1218, 1224, 1230, 1236, 1242, 1248, 1254,
1260, 1266, 1272, 1278, 1284, 1290, 1296, 1302, 1308)

1. which ones are Sums of 2 squares ?

2. while doing these tests by hand, what are some short-cuts ?
(ways of avoiding exhaustive searches)

omg... i almost forgot to ask...
------- pls PLEASE wait a few days before posting answers or hints.

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• From Ilan Mayer@21:1/5 to henh...@gmail.com on Sun Jul 10 09:44:47 2022
On Tuesday, July 5, 2022 at 11:28:59 AM UTC-4, henh...@gmail.com wrote:
Of these numbers ( 2 sets )

Set1= (6, 12, 18, 24, . . . , 150, 156) = (all multiples of 6 under 160)

Set2= (1200, 1206, 1212, 1218, 1224, 1230, 1236, 1242, 1248, 1254,
1260, 1266, 1272, 1278, 1284, 1290, 1296, 1302, 1308)

1. which ones are Sums of 2 squares ?

2. while doing these tests by hand, what are some short-cuts ?
(ways of avoiding exhaustive searches)

omg... i almost forgot to ask...
------- pls PLEASE wait a few days before posting answers or hints.

SPOILER SPACE

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If a square n is not divisible by 3, it has the form (3*m+1)^2 or (3*m+2)^2. In either case this means that n has the form 3*m+1.
A sum of two such squares has the form 3*m+2, and thus is not a multiple of 6. This means the for a multiple of 6 each of the two squares must be the square of a multiple of 3, and both even or both odd.
For set 1 only 9, 36, 81, 144 are candidates.
18 = 9+9, 72 = 36+36 and 90 = 9+81 are the only sums of two squares in this set.
For set 2 only 9, 36, 81, 144, 225, 324, 441, 576, 729, 900, 1089, 1296 are candidates.
1224 = 324+900 is the only sum of two squares in this set.

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__/\\ //\__ Ilan Mayer
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• From henhanna@gmail.com@21:1/5 to Ilan Mayer on Sat Jul 16 07:00:27 2022
On Sunday, July 10, 2022 at 9:44:48 AM UTC-7, Ilan Mayer wrote:
On Tuesday, July 5, 2022 at 11:28:59 AM UTC-4, henh...@gmail.com wrote:
Of these numbers ( 2 sets )

Set1= (6, 12, 18, 24, . . . , 150, 156) = (all multiples of 6 under 160)

Set2= (1200, 1206, 1212, 1218, 1224, 1230, 1236, 1242, 1248, 1254,
1260, 1266, 1272, 1278, 1284, 1290, 1296, 1302, 1308)

1. which ones are Sums of 2 squares ?

2. while doing these tests by hand, what are some short-cuts ?
(ways of avoiding exhaustive searches)

omg... i almost forgot to ask...
------- pls PLEASE wait a few days before posting answers or hints.
SPOILER SPACE

.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
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.
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.
.
.
.
If a square n is not divisible by 3, it has the form (3*m+1)^2 or (3*m+2)^2. In either case this means that n has the form 3*m+1.
A sum of two such squares has the form 3*m+2, and thus is not a multiple of 6.
This means the for a multiple of 6 each of the two squares must be the square of a multiple of 3, and both even or both odd.
For set 1 only 9, 36, 81, 144 are candidates.
18 = 9+9, 72 = 36+36 and 90 = 9+81 are the only sums of two squares in this set.
For set 2 only 9, 36, 81, 144, 225, 324, 441, 576, 729, 900, 1089, 1296 are candidates.
1224 = 324+900 is the only sum of two squares in this set.

For set 2 -- only 9, 36, 81, 144, 225, 324, 441, 576, 729, 900, 1089, 1296 are candidates.
1224 = 324+900 is the only sum of two squares in this set.

Set2= (1200, 1206, 1212, 1218, 1224, 1230, 1236, 1242, 1248, 1254,
1260, 1266, 1272, 1278, 1284, 1290, 1296, 1302, 1308)

i could check (by hand) only a few numbers in Set2 --- is it practical to expect a pretty lazy person to
check all of these by hand ?

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