Find integers a, b such that a+45b, a+5b, a+15b and a-3b are all squares.

Why is this _the_ 4 squares puzzle, seeing that one can ask the same
question for any 4 non-proportional linear forms?  At any rate, such
a problem is always equivalent (up to scaling a and b by squares)
to finding rational points on a certain elliptic curve, and in our
case this elliptic curve has positive rank, so that there are infinitely,
although very widely separated, primitive solutions (i.e. solutions
from which a further square factor cannot be removed).

For the case at hand, let m^2=a+5b and n^2=a-3b; then
a=(3m^2+5n^2)/8 and b=(m^2-n^2)/8 and so a+45b=6m^2-5n^2 and
a+15b=(9m^2-5n^2)/4.  So we're looking for integers m,n such that
6m^2-5n^2 and 9m^2-5n^2 are squares; equivalently, for a rational
number t=m/n such that 6t^2-5 and 9t^2-5 are squares.  The latter
is easy: let 9t^2-5=(3t-r)^2 and find t=(r^2+5)/(6r).  Substituting
this into 6t^2-5 gives (r^4-20r^2+25)/(6r^2), so we're left with
the problem of finding all rational r such that r^4-20r^2+25=6s^2
for some rational s.  We already know one such r, namely r=1
(coming from (a,b)=(1,0), i.e. m=n=t=1).  We could now put this
curve in standard Weierstrass form, and if you want to find all
the solutions you probably have to go through this exercise and
then see for instance the section on explicit 2-descents in
Silverman's _The Arithmetic of Elliptic Curves_.  But if you
just want to obtain further r from the known one and thus exhibit
larger solutions to your original puzzle, there are shortcuts
you can take: for instance, we can ask that
r^4-20r^2+25 [Editor: "r^4-20r^2+5" in the original] = 6(ar^2+br+c)^2
where a,b,c are chosen so that this quartic equation in r has a triple
root at the known r=1, which gives a=-5/3, b=25/3, c=-17/3 and a new
r=-1/23, whence t=-441/23 and (a,b)=(73261,24244).  Repeating this
procedure [and variants -- once you know more solutions (r,s),
you can find parabolas s=ar^2+br+c intersecting
r^4-20r^2+25=6s^2 [Editor: "t^4-20r^2+25=6s^2" in the original]
in any three of them, not necessarily triply at the first one,
and the fourth point of intersection will in general give
a new rational solution] you can obtain ever larger solutions.

-- Noam D. Elkies <elkies@zariski.harvard.edu>

[ Editor: check: with (a,b) = (73261,24244), a+45b = 1164241 = 1079^2,
a+5b = 194481 = 441^2, a+15b = 436921 = 661^2, a-3b = 529 = 23^2. ]