
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 (s3r), (sr), (s+r), (s+3r) are all squares in arithmetic progression, and their common difference, 2r is smaller than 2abcd contradicting minimality.
Suppose there exist four squares A², B², C², D² in increasing arithmetic progression with a minimal common difference, i.e., we have
B²A² = C²B² = D²C²
being the smallest of any set of four squares in arithmetic progression. We can assume the squares are mutually coprime, 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 coprime integers u,v such that
u=(C+A)/2,
v=(CA)/2so then,
A=uv,
C=u+v,
u²+v²=B²,
and the common difference of the progression is (C²A²)/2=2uv.
We also have
D²  B² = 4uv,
which factors as
[(D+B)/2][(DB)/2] = uv.
The two factors on the left are coprime, as are u and v, so there exist four mutually coprime integers a,b,c,d (exactly one even) such that
u=ab,
v=cd,
D+B=2ac, and
DB=2bd.
This implies B=acbd, so we can substitute into the equation u²+v²=B² to give
(ab)² + (cd)² = (acbd)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(a^{4}  a²d² + d^{4})) / (a²  d²)
From this quadratic equation we find that c is a rational function of sqrt(a^{4}  a²d² + d^{4}), which implies there is an odd integer m such that
a^{4}  a²d² + d^{4} = m².
(I will refer to this later as the "quartic equation")
Since a and d are odd there exist coprime integers x and y (exactly one even) such that
a² = k(x+y) and
d² = k(xy), where k=±1.
Substituting into the above quartic gives
(k(x+y))²  k(x+y)k(xy) + (k(xy))² = m²
x² + 2xy + y²  x² + y² + x²  2xy + y² = m²
x² + 3y² = m²
3y² = m²x²
from which it's clear (using mod 4 parity) that y must be even and x odd. Since (m+x)(mx) is divisible by 3, we know that either m+x is divisible by 3 or mx 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][(mx)/2],
which implies that (m+x)/2 is three times a square, and (mx)/2 is a square. Thus we have coprime integers r and s (one odd and one even) such that
(m+x)/2=3r²,
(mx)/2=s²,
m=3r²+s²,
x=3r²s²
From this, we can find y:
y²=(m²x²)/3
y²=((3r²+s²)²(3r²s²)²)/3
y²=12r²s²/3
y=±2rs.
Substituting for x and y back into the expressions for a²=k(x+y) and d²=k(xy) (and transposing a and d if necessary, since the quartic is symmetrical with respect to them) gives
a²=k(3r²s²+2rs) and
d²=k(3r²s²2rs).a²=k(s+r)(s3r) and
d²=k(sr)(s+3r).
Since the right hand factors are coprime, it follows that the four quantities (s3r), (sr), (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=11 (mod 4), which is false.
Therefore, we must have 3r < s, so from m=3r²+s² we have
12r² < m.
Also the quartic equation
a^{4}  a²d² + d^{4} = m²
a^{4} + 2a²d² + d^{4} > 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 (2x)(2y) = 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 2b²a² and c² is coprime to 2d²c² 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
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)
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.