# n^4 + 4*m^4 = p

FabianFnas
Posers and Puzzles 05 Nov '08 17:25
1. 05 Nov '08 17:25
Consider this equation:
n^4 + 4*m^4 = p
where x and y is an positive integer and p is a prime.

Does this equation have
(a) no solutions,
(b) one solution,
(c) many solutions, or
(d) infinitely many solutions?

If possible, show the a, b and p in the solution(s).
2. 05 Nov '08 17:48
Originally posted by FabianFnas
Consider this equation:
n^4 + 4*m^4 = p
where x and y is an positive integer and p is a prime.

Does this equation have
(a) no solutions,
(b) one solution,
(c) many solutions, or
(d) infinitely many solutions?

If possible, show the a, b and p in the solution(s).
perhaps a silly question... but 3 pairs of variables? n-m x-y or a-b? i assume you mean to have the question be consistent in choosing one of these pairings (i.e. n^3 + 4*m^4 ... where n and m are pos int.... find n and m)?

in terms of finding the solution, i have to think for a bit ðŸ™‚
3. Palynka
Upward Spiral
05 Nov '08 18:221 edit
One solution: (m,n,p) = (1,1,5)

There aren't any other ones. Now prove me wrong. ðŸ˜›
4. 05 Nov '08 20:253 edits
Couldn't (-1, -1, 5) be correct?

-1^4=1, 1+(4*1)=5

Oh, and ðŸ˜›
5. 05 Nov '08 22:08
Originally posted by ketch90
Couldn't (-1, -1, 5) be correct?

-1^4=1, 1+(4*1)=5

Oh, and ðŸ˜›
No, -1 is negative integer, not positive integer.

And right - misprint. Should be a^4 + 4*b^4 = p.

Okay, we have one solution. So the answer of (a) is no, and proven. One solution is found.
But are there others? Yes? Give examples! No? Prove it!
Are there infinitely many? Prove it!
6. 06 Nov '08 00:51
Any other solution will require either a or b be divisible by 5, and a be odd.

The reason being any number not divisible by 5 will have a 4th power of the form 5k+1, so if you add 5 of these together (4 b^4s and a a^4a), the answer will be divisible by 5.

So here's what I have so far as being necessary for any solution.

* a is odd, otherwise p is divisible by 4.
* a and b are relatively prime, obviously..
* either a or b is divisible by 5, save for the singular counterexample of a=1, b=1
7. sonhouse
Fast and Curious
06 Nov '08 10:221 edit
Originally posted by Palynka
One solution: (m,n,p) = (1,1,5)

There aren't any other ones. Now prove me wrong. ðŸ˜›
Why isn't P=257 if m and n is 1? 1^4+ 4*1^4=257
Oh, I think the equation should have been written
X^4+ 4*(Y^4)
That makes it =5. Otherwise its 257
I took it to be 1^4 + (4*1)^4 =P
8. Palynka
Upward Spiral
06 Nov '08 12:23
Originally posted by sonhouse
Why isn't P=257 if m and n is 1? 1^4+ 4*1^4=257
Oh, I think the equation should have been written
X^4+ 4*(Y^4)
That makes it =5. Otherwise its 257
I took it to be 1^4 + (4*1)^4 =P
PEMDAS.
9. ChronicLeaky
Don't Fear Me
06 Nov '08 21:081 edit
I'd keep the things geepamoogle mentioned in mind and then try factorising b^4 over the ring of integers in Q[sqrt(p)], ie the ring D = Z[(1+sqrt(p))/2]. This gives, where d = (1+sqrt(p))/2:

4b^4 = (2d -a^2 -1)(2d + a^2 - 1)

Now let (a,b) be an integer solution for some given prime p and consider the ideals H = (2d - a^2 -1) and I = (2d + a^2 -1) in the ring where we live. Then there's some business about checking the ideal class group of Q[sqrt(p)]. You ultimately get some other equation whose integral solutions give rise to the integral solutions of the equation you want. I forget the details right now, but I can refresh my memory later perhaps.

EDIT On second thought, it isn't at all easy to see what the ideal class group will look like in general, so this probably isn't the approach to take.
10. 07 Nov '08 10:221 edit
Hint:
See if a factorization of a^4 + 4*b^4 gives something...

Did 4b^4 = (2d -a^2 -1)(2d + a^2 - 1), where d = (1+sqrt(p))/2, give something? I'd rather try a^4 + 4*b^4 and care about p later...

There can be, of course, several lines to the solution. I have one of them, others I cannot perhaps even understand...
11. ChronicLeaky
Don't Fear Me
07 Nov '08 13:071 edit
Oh. Now I feel silly for talking about class groups, because:

(a^4 + 4b^4) = (a^2 + 2ab + 2b^2)(a^2 - 2ab + 2b^2)

If this is to be prime, then the smaller of the two factors must be 1. Now either a = 1 = b or:

a^2 - 2ab + 2b^2 = (a-b)^2 + b^2 = 1, so the above is the only solution.
12. ChronicLeaky
Don't Fear Me
07 Nov '08 13:11
Originally posted by Palynka
PEMDAS.
Palynka Expresses Mathematical Doctrine Acronym-Style?
13. Palynka
Upward Spiral
07 Nov '08 14:11
Originally posted by ChronicLeaky
Palynka Expresses Mathematical Doctrine Acronym-Style?
Palynka Exhibits Mortified Dermis in Avatar Snip
14. 07 Nov '08 14:31
Originally posted by ChronicLeaky
Oh. Now I feel silly for talking about class groups, because:

(a^4 + 4b^4) = (a^2 + 2ab + 2b^2)(a^2 - 2ab + 2b^2)

If this is to be prime, then the smaller of the two factors must be 1. Now either a = 1 = b or:

a^2 - 2ab + 2b^2 = (a-b)^2 + b^2 = 1, so the above is the only solution.
Nice done!
15. 07 Nov '08 15:00
I didn't think an expression like a^4 + 4*b^4 would factor into 2 parts so nicely.

Nice job..

Just as a small note to an above post, standard order of operations dictates that all powers operations (squares, cubes, etc) are evaluated before multiplication and division.

http://en.wikipedia.org/wiki/Order_of_operations

The order, as I understand it.

1) Expressions inside parentheses
2) Exponents/Roots
3) Multiplcation/Division