8k+7 = 2x^2 + y^2 + z^2

[Shiv K. Gupta]

Dear Number Theorists,

While discussing the representation of integers as sums of squares in an
 
introductory course on number theory a student observed that any number

which is 7 mod 8, can always be expressed as 2x^2 + y^2 + z^2. It appears
 
to be true. Can anyone suggest an elementary  proof?
- Shiv Gupta
West Chester University

[zwsun]

Dear number theorists,

  Shiv K. Gupta  asked for a simple proof of the observation that 8k+7
can always be expressed as 2x^2 + y^2 + z^2.

  In fact ,if we write 4n+2=(2x)^2+y^2+z^2 with y=z(mod 2)
Then 2n+1 =2x^2+(y^2+z^2)/2  = 2x^2+((x+y)/2)^2+((x-y)/2)^2.
So,the Gauss-Legendre theorem on sums of three squares implies that each
positive odd integer can be written in the form of 2x^2+y^2+z^2.
This was first observed by Euler and rediscovered by L.Panaitopol [Amer.
Math, Monthly 112(2005),168-171)]  who showed that each positive odd
integer  has the form  ax^2+by^2+cz^2 (0<a \le b \le c) if and only if
the vector (a,b,c) is (1,1,2) or (1,2,3) or (1,2,4).

  Motivated by Panaitopol's work, in 2005 I began to investigate  for
what kind of positive integers a,b,c each natural number can be written
in the form ax^2+by^2+cT_z or ax^2+bT_y+cT_z,where T_m denotes the
triangular number m(m+1)/2. This project was completed in a series of
papers with the title "Mixed sums of squares and triangular numbers".

  Sincerely,

Zhi-Wei Sun
http://math.nju.edu.cn/~zwsun

[Everett W. Howe]

Shiv Gupta asks:

> While discussing the representation of integers as sums of squares  
> in an
> introductory course on number theory a student observed that any  
> number
> which is 7 mod 8, can always be expressed as 2x^2 + y^2 + z^2. It  
> appears
> to be true. Can anyone suggest an elementary  proof?


If you've already proven the three-squares theorem [that the positive  
integers of the form x^2 + y^2 + z^2 are precisely the positive  
integers not of the form 4^k*(8*n+7) ] then your student's observation  
follows easily:

Given an odd number n, write 2*n as the sum of three squares, say
   2*n = x^2 + y^2 + z^2.
Since 2*n is congruent to 2 mod 4, exactly two of the integers x, y,  
and z must be odd; say that x and y are odd and z is even.  Then set
   X = (x+y)/2
   Y = (x-y)/2
   Z = z/2.
It follows that X^2 + Y^2 + 2*Z^2 = n.

I got this argument from Dickson (Integers represented by positive  
ternary quadratic forms, Bull. Amer. Math. Soc 33 (1927) pp. 63--70),  
who shows that in fact the positive integers represented by x^2 + y^2  
+ 2*z^2 are precisely the positive integers not of the form 4^k*(16*n  
+ 14).

Dickson's paper is available from Project Euclid at this URL:
   http://projecteuclid.org/euclid.bams/1183491956

-- Everett



________________________________________________________________________
Everett Howe                          Center for Communications Research
however@...                           4320 Westerra Court
http://www.alumni.caltech.edu/~however/             San Diego, CA  92121

[Schroeppel, Richard]

 Suppose N is 7 mod 8.
Then 2N is 6 mod 8, and (by the 3 square thm) has a sum-of-3-squares representation.

2N = A^2 + B^2 + C^2

Looking at the situation mod 8, we may assume A is even, and B and C are odd.

Then  N =  2 (A/2)^2 + ((B+C)/2)^2 + ((B-C)/2)^2.

A kludge, but elementary.

Rich Schroeppel

-----Original Message-----
From: Number Theory List [mailto:NMBRTHRY@...] On Behalf Of Shiv K. Gupta
Sent: Friday, May 16, 2008 10:30 AM
To: NMBRTHRY@...
Subject: 8k+7 = 2x^2 + y^2 + z^2

Dear Number Theorists,

While discussing the representation of integers as sums of squares in an

introductory course on number theory a student observed that any number

which is 7 mod 8, can always be expressed as 2x^2 + y^2 + z^2. It appears

to be true. Can anyone suggest an elementary  proof?
- Shiv Gupta
West Chester University

[Leonard Bailey]

Hi,
In Exercise 4 on p395 of Elementary Theory of Numbers by W. Sierpinski

(Ed. A. Schinzel) North-Holland 1988, the author proves that any odd
natural number 2n + 1 = a^2 + b^2 + 2c^2. He uses a theorem first prove
d
by Gauss to the effect that a natural number can be the sum of three
squares iff it is NOT of the form 4^m(8k + 7) where m and k are non-
negative integers, the full proof of which is not given in the above
reference.
The a, b, c above are integers, in effect 0 or natural numbers, for examp
le
3 = 0^2 + 1^2 + 2*1^2.
The numbers 8k + 7 therefore can likewise be expressed.

There remains the issue of whether in this restricted case the a, b, c ar
e
always natural numbers (that is excluding 0).
The answer is yes because 8k + 7 cannot be a perfect square, nor the sum

of two squares, nor can it take the form b^2 + 2c^2. This last can be
shown by substituting for b and c two evens, two odds, even and odd,
odd and even, to obtain forms that are not 8k + 7.

Regards, Leonard Bailey