September 10th, 2019 at 2:55:32 PM
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I have noticed the following:

No integer congruent with 7(mod 8) is ever a factor/divisor of any number n

For example, 7, 15 and 23 are each congruent with 7(mod 8) but they are not a factor/divisor of any value of n

I have been fooling around with modular math, and its pretty clear that for all possible values of n that n

So, I'm stuck. I don't know enough modular arithmetic, I guess. Can anyone help with this?

No integer congruent with 7(mod 8) is ever a factor/divisor of any number n

^{2}+2.For example, 7, 15 and 23 are each congruent with 7(mod 8) but they are not a factor/divisor of any value of n

^{2}+2: 3, 6, 11, 18, 27, 38, 51, 66, 83, 102, 123, 146, 171, 198, 227, 258, 291 . . . . 4016018, 4020027 . . .I have been fooling around with modular math, and its pretty clear that for all possible values of n that n

^{2}+2 is always congruent with 2,3 or 6 (mod 8). But I can't see the reason why a number that is 7(mod 8) -which also can be written as -1(mod 8) - can't be a factor/divisor of a number that is 2,3 or 6 (mod 8)So, I'm stuck. I don't know enough modular arithmetic, I guess. Can anyone help with this?

So many better men, a few of them friends, were dead. And a thousand thousand slimy things lived on, and so did I.

September 10th, 2019 at 3:08:14 PM
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Not thought about your question, but it reminds me of a result I encountered recently:

All integers are the sum of three squares, except those congruent to 7(mod 8).

Maybe a link?

All integers are the sum of three squares, except those congruent to 7(mod 8).

Maybe a link?

Reperiet qui quaesiverit

September 10th, 2019 at 4:54:12 PM
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Yes,you remind me that I've read that 7(mod 8) integers are the sum of four squares. Weird.

So many better men, a few of them friends, were dead. And a thousand thousand slimy things lived on, and so did I.

September 11th, 2019 at 5:19:08 AM
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As n

But that doesn’t say anything about its factors. For exemple, 2*(8+7)=30=25+4+1 : m*(8x+7) must be the sum of three squares if m is not 4

Are you sure about the original assertion or is it a conjecture?

^{2}+2 = n^{2}+1+1 is the sum of three squares, it cannot be =7(mod8).But that doesn’t say anything about its factors. For exemple, 2*(8+7)=30=25+4+1 : m*(8x+7) must be the sum of three squares if m is not 4

^{k}(and cannot be if m= 4^{k}).Are you sure about the original assertion or is it a conjecture?

Last edited by: kubikulann on Sep 11, 2019

Reperiet qui quaesiverit

September 11th, 2019 at 6:43:37 AM
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Looking for an integer n such that n

Defining n = 8u+v we have the following constraints modulo 8:

64u

Or, modulo 8,

Alas, these are necessary Conditions, not sufficient. For ex., third line v=2, a=1 b=1 u=2 fills the modulo condition, but it gives n=18 -> n

What is left to do is:

Gather all triplets (u,a,b) respecting the necessary condition ;

Then

either prove that it is impossible to respect the 2d table condition,

Or find an example, which makes the original assumption false.

^{2}+2 equals (8a+7)(8b+c).Defining n = 8u+v we have the following constraints modulo 8:

v | n^{2}+2 | c |
---|---|---|

0,4 | 2 | 6 |

1,3,5,7 | 3 | 5 |

2,6 | 6 | 2 |

64u

^{2}+16uv+v^{2}+2 = 64ab+8(ac+7b)+7cv | v^{2}+2 | 7c | diff | 7b+ac=8(u^{2}-ab)+2uv+diff/8 |
---|---|---|---|---|

0 | 2 | 42 | -40 | 8(u^{2}-ab)-5 |

1 | 3 | 35 | -32 | 8(u^{2}-ab)+2u-4 |

2 | 6 | 14 | 8 | 8(u^{2}-ab)+4u+1 |

3 | 11 | 35 | 24 | 8(u^{2}-ab)+6u+3 |

4 | 18 | 42 | 24 | 8(u^{2}-ab)+8u+3 |

5 | 27 | 35 | 8 | 8(u^{2}-ab)+10u+1 |

6 | 38 | 14 | -24 | 8(u^{2}-ab)+12u-3 |

7 | 51 | 35 | -16 | 8(u^{2}-ab)+14u-2 |

Or, modulo 8,

v | 7b+ac | = |
---|---|---|

0 | 7b+6a | -5 |

1 | 7b+5a | 2u-4 |

2 | 7b+2a | 4u+1 |

3 | 7b+5a | 6u+3 |

4 | 7b+6a | +3 |

5 | 7b+5a | 2u+1 |

6 | 7b+2a | 4u-3 |

7 | 7b+5a | 6u-2 |

Alas, these are necessary Conditions, not sufficient. For ex., third line v=2, a=1 b=1 u=2 fills the modulo condition, but it gives n=18 -> n

^{2}+2=326, which is not equal to (15)*(10).What is left to do is:

Gather all triplets (u,a,b) respecting the necessary condition ;

Then

either prove that it is impossible to respect the 2d table condition,

Or find an example, which makes the original assumption false.

Reperiet qui quaesiverit

September 11th, 2019 at 12:00:32 PM
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Quote:kubikulannAs n

^{2}+2 = n^{2}+1+1 is the sum of three squares, it cannot be =7(mod8).

But that doesn’t say anything about its factors. For exemple, 2*(8+7)=30=25+4+1 : m*(8x+7) must be the sum of three squares if m is not 4^{k}(and cannot be if m= 4^{k}).

Are you sure about the original assertion or is it a conjecture?

It is a conjecture, based on >20,000 trials.

Thanks for looking at this, I am reading everything you post.

Last edited by: gordonm888 on Sep 11, 2019

So many better men, a few of them friends, were dead. And a thousand thousand slimy things lived on, and so did I.

September 11th, 2019 at 12:49:08 PM
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Quote:kubikulannLooking for an integer n such that n

^{2}+2 equals (8a+7)(8b+c).

Defining n = 8u+v we have the following constraints modulo 8:

v n ^{2}+2c 0,4 2 6 1,3,5,7 3 5 2,6 6 2

64u^{2}+16uv+v^{2}+2 = 64ab+8(ac+7b)+7c

v v ^{2}+27c diff 7b+ac=8(u ^{2}-ab)+2uv+diff/80 2 42 -40 8(u ^{2}-ab)-51 3 35 -32 8(u ^{2}-ab)+2u-42 6 14 8 8(u ^{2}-ab)+4u+13 11 35 24 8(u ^{2}-ab)+6u+34 18 42 24 8(u ^{2}-ab)+8u+35 27 35 8 8(u ^{2}-ab)+10u+16 38 14 -24 8(u ^{2}-ab)+12u-37 51 35 -16 8(u ^{2}-ab)+14u-2

Or, modulo 8,

v 7b+ac = 0 7b+6a -5 1 7b+5a 2u-4 2 7b+2a 4u+1 3 7b+5a 6u+3 4 7b+6a +3 5 7b+5a 2u+1 6 7b+2a 4u-3 7 7b+5a 6u-2

Alas, these are necessary Conditions, not sufficient. For ex., third line v=2, a=1 b=1 u=2 fills the modulo condition, but it gives n=18 -> n^{2}+2=326, which is not equal to (15)*(10).

What is left to do is:

Gather all triplets (u,a,b) respecting the necessary condition ;

Then

either prove that it is impossible to respect the 2d table condition,

Or find an example, which makes the original assumption false.

Wow! Wow! I am VERY impressed. Okay, my head is exploding, but after studying it, I have followed this through and understand the point you are at. For each of the eight possible pairs of (v, c) as defined by the 1st table, it is necessary to define triplets (u,a,b) that satisfy the v-dependent modulo conditions of the last table and either find an example for which (8u+v)

^{2}+2 = (8a+7)*(8b+c) or show algebraicly that it can't be solved.

Are a and b allowed to be negative? If not, then there is no possible (u,a,b) triplet that satisfies the modulo conditions for v=0. Or for v=4.

EDIT: My general impression is that, always, u>b and u>a. Therefore, it seems that (8u+v)

^{2}is always greater than (8a+7)*(8b+c).

But why? If we had done this for 1,3,5(mod 8) we should find that there are plenty of solutions. What is there about 7mod(8) that makes it have zero solutions? And what is there about n

^{2}+2? Because 7(mod8) numbers divide into n

^{2}+3. Example: 7*12=84=9

^{2}+3

Last edited by: gordonm888 on Sep 11, 2019

September 11th, 2019 at 3:55:20 PM
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Quote:Are a and b allowed to be negative? If not, then there is no possible (u,a,b) triplet that satisfies the modulo conditions for v=0. Or for v=4.

I thought so (even wrote it down, but later edited).

Actually, as it is modular, it doesn’t matter if a and b are positive or not.

Writing [a] for a(mod8),

what we have for v=1 is

7[a]+5[b]=2[u]-4

Example

For v=0 : 7[a]+6[b]+5=0 (mod 8)

Solutions : ([a],[b]) = (1,2),(1,6),(3,1),(3,5),(5,0),(5,4),(7,3),(7,7)

(8a+6)(8b+7) = 14*23,14*55,30*15,30*47,46*7,46*39,62*31,62*63

(8a+6)(8b+7)-2 = 320, 768 , 448 , 1408 , 320 , 1792 , 1920 , 3904

none of which is a square.

But there are many other solutions to (a,b) ; (9,2),(9,6),(17,2),(17,6)…(1,10),(1,18),(9,10),(9,18)… ad inf

The equation becomes

u

^{2}= (8(a+8i)+6)(8(b+8j)+7)-2

= (8a+6)(8b+7)-2 + 64i(8b+7) + 64j(8a+6) + 64

^{2}ij

Ouch!

The underlined part mod 64 can be 0,0,0,0,0,0,0,0

so we are looking for u such that u

^{2}=0(mod64)

This is clearly not impossible.

One more turn on the merry-go-round.

( u/8 )

^{2}= <5,12,7,22,5,28,30,61> +i(8b+7) + j(8a+6) + 64ij

= <5,12,7,22,5,28,30,61> + <23,55,15,47,7,39,31,63>i + <14,14,30,30,46,46,62,62>j + 64ij

Eight equations with the question, does there exist i and j such that it is a square?

(It.s 1AM now. To be followed tomorrow.)

Reperiet qui quaesiverit

September 11th, 2019 at 5:51:06 PM
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Quote:gordonm888Quote:kubikulannAre you sure about the original assertion or is it a conjecture?

It is a conjecture, based on >20,000 trials.

I started a brute force search, and could not find any n < 16,000 such that n

^{2}+ 2 was a multiple of any number of the form 8k + 7.

EDIT: None found through 25,000

I have also asked the math boffins at Art of Problem Solving to see if they can come up with anything.

EDIT: Somebody has come up with a solution (that confirms the conjecture) that I need to check; it involves "quadratic residues" and "Legendre symbols"

Last edited by: ThatDonGuy on Sep 11, 2019

September 11th, 2019 at 7:51:12 PM
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I have been checking the literature, mainly by mining the On-line Encyclopedia of Integer Sequences, oeis.org

Sequence A059100 is a(n)=n^2+2. There is not any mention of the numbers in this sequence never being a multiple of 8k+7.

Numbers of the form nSo many better men, a few of them friends, were dead. And a thousand thousand slimy things lived on, and so did I.

Sequence A059100 is a(n)=n^2+2. There is not any mention of the numbers in this sequence never being a multiple of 8k+7.

Numbers of the form n

^{2}+m^{2}, such as n^{2}+1, n^{2}+4, n^{2}+9, etc. are the sums of squares and are known to be divisible only by integers congruent to 1(mod 4); that is, divisible by integers congruent 1,5(mod 8). However, I have not seen any mention of other values of (n^{2}+/- m) having any limits on factors/divisors.