Arithmetic Sequences of Perfect Squares

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Arithmetic Sequences of Perfect Squares

{1, 25, 49} is a 3-element arithmetic progression of perfect squares. Are there any 4- or more-element arithmetic progressions of perfect squares?

If {a², b², c², d²} is such a sequence, then

a²+d² = b²+c²

If we let

n = (c-a)/2, and
m = (c+a)/2

as we did on the previous page, then

a² = (m-n)² = m² - 2mn + n²
b² = m² + n²
c² = (m+n)² = m² + 2mn + n²
d² = m² + 4mn + n²

It was pointed out that a²+d² = b²+c², so the following statements are also true:

Let A = a²+d²,
let B = b²+c²

Since A=B, AB is a perfect square.

AB = (a²+d²)(b²+c²) = (ab-dc)²+(ac+bd)² = (ab+dc)²+(ac-bd)²

AB = ((m-n)b-d(m+n))²+((m-n)(m+n)+bd)²

Does this get me anywhere? I don't know...

Here's a different approach...

Remember, we generated all the Pythagorean Triples from s and r, where s and r are coprime positive integers of opposite parity.

The primitive pythagorean triple n, m, b (n²+m²=b²) is generated by

n = r²-s²
m = 2rs
b = r²+s²

Then we generated the arithmetic sequence of squares by

a²	= (m-n)²	= (-r² + 2rs + s²)²	= r⁴ - 4r³s + 2r²s² + 4rs³ + s⁴
b²		= (r² + s²)²	= r⁴ + 2r²s² + s⁴
c²	= (m+n)²	= (r² + 2rs - s²)²	= r⁴ + 4r³s + 2r²s² - 4rs³ + s⁴
Then, by extension,
d²			= r⁴ + 8r³s + 2r²s² - 8rs³ + s⁴

The trick would be to show that this expression can't be a perfect square

r⁴ + 8r³s + 2r²s² - 8rs³ + s⁴

= r⁴ + 4r³s + 2r²s² - 4rs³ + s⁴ + 4r³s - 4rs³

= (r² + 2rs - s²)² + 4(r)(s)(r-s)(r+s)

Once again, I'm out of ideas...

A search of the internet turned up http://www.mathpages.com/home/kmath044.htm which gives a proof:

No Four Squares In Arithmetic Progression

Dickson's "History of the Theory of Numbers" presents a proof that there do not exist four squares in arithmetic progression, but I'm not very happy with it. I've also received email from various people saying they didn't understand the proof. Everyone seems to agree that the difficulty with the proof (which Dickson attributes to Bronwin and Furnass) is at this point:

      ... from the relations

                 y² - x²  =  z² - y²  =  w² - z² 

  we know there are integers a,b,c,d such that
 
                     y+x = 2ab     y-x = 2cd     
                     z+y = 2ac     z-y = 2bd
                     w+z = 2bc     w-z = 2ad

I really don't see why there have to be four integers satisfying all six of these relations. The rest of the proof relies entirely on this step. I'd be interested if anyone could explain why all six of these relations must be satisfied by four integers a,b,c,d. Maybe someone with access to a reference library could check Dickson's sources to be sure he got it right.

Anyway, after giving up on this approach, I decided to try proving the proposition on my own. My proof starts with A², B², C², D² in arithmetic progression with minimal common differences. It works by finding 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 I get from the arithmetic progression I 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. Here's what I came up with:

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 have co-prime integers u,v such that

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.

This quadratic is symmetrical in the four variables, 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

         2b² - a²      2d² - c²
         ---------     ----------   =   1
             b²            d²

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

         b² - d² = d² - c²         d² - b² = b² - a²

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

Related pages in this website

Arithmetic Sequence of Perfect Squares

Number Theory

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