Number Theory 677So 3 dividesz^2 +t^2. Since the residue of a square modulo 3 is either 0 or 1, this can
happen only if bothzandtare divisible by 3, meaning thatz= 3 z 1 ,t= 3 t 1. But then
6 (x^2 +y^2 )= 9 (z^21 +t 12 ),and hencex^2 +y^2 is divisible by 3. Again, this can happen only ifx = 3 x 1 , and
y= 3 y 1 , withx 1 ,y 1 positive integers. So(x 1 ,y 1 ,z 1 ,t 1 )is another solution. We construct
inductively a decreasing infinite sequence of positive solutions, which, of course, cannot
exist. Hence the system does not admit nontrivial solutions.
(W. Sierpinski, 250 ́ Problems in Elementary Number Theory,Pa ́nstwowe Wydaw-
nictwo Naukowe, Warsawa, 1970)
708.Assume that the positive integersx, y, zsatisfy the given equation, and letd=xy.
Ifd=1, thenx=y=1 andz=0, which cannot happen. Henced>1. Letpbe a
prime divisor ofd. Because
(x+y)(x−y)=x^2 −y^2 = 2 xyz≡ 0 (modp),eitherx≡y(modp)orx≡−y(modp). Butpdivides one ofxandy,sopmust divide
the other, too. Hencex 1 =x/pandy 1 =y/pare positive integers, andx 1 ,y 1 ,zsatisfy
the given equation as well. Repeating the argument, we construct an infinite sequence of
solutions(xn,yn,z),n≥1, to the original equation, withx 1 >x 2 >x 3 >···.This is,
of course, impossible; hence the equation has no solutions.
(T. Andreescu, D. Andrica,An Introduction to Diophantine Equations, GIL, 2002)
709.If(an^2 )nis an infinite arithmetic progression, then
ak^2 + 1 −ak^2 =a^2 k−ak^2 − 1 , fork≥ 2.Such an arithmetic progression must be increasing, soak+ 1 +ak>ak+ak− 1. Combining
the two relations, we obtainak+ 1 −ak<ak−ak− 1 , for allk≥2. We have thus obtained
an infinite descending sequence of positive integers
a 2 −a 1 >a 3 −a 2 >a 4 −a 3 >···.Clearly, such a sequence cannot exist. Hence there is no infinite arithmetic progression
whose terms are perfect squares.
Remark.In fact, much more is true. No four perfect squares can form an arithmetic
progression.
(T.B. Soulami,Les olympiades de mathématiques: Réflexes et stratégies, Ellipses,
1999)