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 Skip Navigation LinksMath Help > Number Theory > Squares > 4 Squares in Sequence > 4 Squares in Sequence, pg 3

Arithmetic Sequence of Four Squares

Proof 1: No Four Squares In Arithmetic Progression

Overview

My objective in presenting this proof is to make it understandable to someone without much formal training in Number Theory.  So I explain everything albeit tersely, so that you will be able to follow the proof.

This is a proof by infinite descent.  We start with A², B², C², D² in arithmetic progression with minimal common differences.  Then we find four integers a,b,c,d such that 2abcd is the common difference in the arithmetic progression.  Then, from some facts about these integers that we get from the arithmetic progression we find two other integers, s,r, such that (s-3r), (s-r), (s+r), (s+3r) are all squares in arithmetic progression, and their common difference, |2r| is smaller than |2abcd| contradicting minimality.

Proof 1: No Four Squares In Arithmetic Progression

Suppose there exist four squares A², B², C², D² in increasing arithmetic progression with a minimal common difference, i.e., we have

B²-A² = C²-B² = D²-C²

being the smallest of any set of four squares in arithmetic progression. We can assume the squares are mutually co-prime, and the parity (mod 4) of the equation shows that either each square must be odd or each square must be even.  But since they're coprime they must all be odd.  Then we can find co-prime integers u,v such that

u=(C+A)/2,
v=(C-A)/2

so then,

A=u-v,
C=u+v,
u²+v²=B²,

and the common difference of the progression is (C²-A²)/2=2uv.

We also have

D² - B² = 4uv,

which factors as

[(D+B)/2][(D-B)/2] = uv.

The two factors on the left are co-prime, as are u and v, so there exist four mutually co-prime integers a,b,c,d (exactly one even) such that

u=ab,
v=cd,
D+B=2ac, and
D-B=2bd.

This implies B=ac-bd, so we can substitute into the equation u²+v²=B² to give

(ab)² + (cd)² = (ac-bd)²
(ab)² + (cd)² - (ac)² - (bd)² + 2abcd = 0

This quadratic is symmetrical in the four variables (that is to say if any of the permutations cdab, badc, or dcba are used in place of abcd the equations, above, are still true), so we can assume c is even and a,b,d are odd.  We can solve for c, giving

c = (abd ± b sqrt(a4 - a²d² + d4)) / (a² - d²)

From this quadratic equation we find that c is a rational function of sqrt(a4 - a²d² + d4), which implies there is an odd integer m such that

a4 - a²d² + d4 = m².
(I will refer to this later as the "quartic equation")

Since a and d are odd there exist co-prime integers x and y (exactly one even) such that

a² = k(x+y) and
d² = k(x-y), where k=±1.

Substituting into the above quartic gives

(k(x+y))² - k(x+y)k(x-y) + (k(x-y))² = m²
x² + 2xy + y² - x² + y² + x² - 2xy + y² = m²
x² + 3y² = m²
3y² = m²-x²

from which it's clear (using mod 4 parity) that y must be even and x odd.  Since (m+x)(m-x) is divisible by 3, we know that either m+x is divisible by 3 or m-x is divisible by 3.  If m+x is not divisible by 3, we could have picked different numbers x,y,k, above, by changing each of their signs.  So m+x is divisible by 3.

3(y/2)² = [(m+x)/2][(m-x)/2],

which implies that (m+x)/2 is three times a square, and (m-x)/2 is a square. Thus we have co-prime integers r and s (one odd and one even) such that

(m+x)/2=3r²,
(m-x)/2=s²,
m=3r²+s²,
x=3r²-s²

From this, we can find y:

y²=(m²-x²)/3
y²=((3r²+s²)²-(3r²-s²)²)/3
y²=12r²s²/3
y=±2rs.

Substituting for x and y back into the expressions for a²=k(x+y) and d²=k(x-y) (and transposing a and d if necessary, since the quartic is symmetrical with respect to them) gives

a²=k(3r²-s²+2rs) and
d²=k(3r²-s²-2rs).

a²=k(s+r)(s-3r) and
d²=k(s-r)(s+3r).

Since the right hand factors are co-prime, it follows that the four quantities (s-3r), (s-r), (s+r), (s+3r) must each have square absolute values, and they are in arithmetic progression with a common difference of 2r. These quantities must all have the same sign, because otherwise the sum of two odd squares would equal the difference of two odd squares, i.e., 1+1=1-1 (mod 4), which is false.

Therefore, we must have |3r| < s, so from m=3r²+s² we have

12r² < m.

Also the quartic equation

a4 - a²d² + d4 = m²
a4 + 2a²d² + d4 > m²
(a² + d²)² > m²
m < a² + d²
12r² < a² + d²
4r² < (a² + d²)/3
4r² < (2 max(a,d)²)/3
|2r| < |sqrt(2/3) max(a,d)|

Thus we have four squares in arithmetic progression with the common difference |2r| < |2abcd|, the latter being the common difference of the original four squares. This contradicts the fact that there must be a smallest absolute common difference for four squares in arithmetic progression.


This theorem can be used to answer other questions as well. For example, H. Jurjus asked whether there are rational numbers p, q such that (p², q²) is a point on the hyperbola given by (2-x)(2-y) = 1 with (p², q²) not equal to (1,1). The answer is no. To see why, suppose p=a/b and q=c/d (both fractions reduced to least terms). Then if (p²,q²) is on the hyperbola we have

         2b� - a�      2d� - c�
         ---------     ----------   =   1
             b�            d�

Since b² is coprime to 2b²-a² and c² is coprime to 2d²-c² it follows that b² = 2d² - c² and d² = 2b² - a². Rearranging terms give

         b� - d� = d� - c�         d� - b� = b� - a�

Together these equations imply that a², b², d² and c² are in arithmetic progression, which is impossible.

Well, that was Proof 1 -- now, take a look at Proof 2

Internet references

Kevin Brown's Mathpages: www.mathpages.com/home/kmath044/kmath044.htm provided this proof that there aren't four squares in arithmetic progression. (index of mathpages)

Related Pages in this website

Arithmetic Sequence of Squares -- a series of rambling discussions about the possible existence of an arithmetic sequence of four squares

 

The webmaster and author of this Math Help site is Graeme McRae.