Number Theory 721814.Divide through byx^2 y^2 to obtain the equivalent equation
1
y^2+
1
xy+
1
x^2= 1.
One of the denominators must be less than or equal to 3. The situationsx=1 andy= 1
are ruled out. Thus we can have onlyxy=2 or 3. But then again eitherxoryis 1,
which is impossible. Hence the equation has no solutions.
815.Note that 2002= 34 + 54 + 64. It suffices to consider
xk= 3 · 2002 k,yk= 5 · 2002 k,zk= 6 · 2002 k,wk= 4 k+ 1 ,withka positive integer. Indeed,
xk^4 +yk^4 +z^4 k=( 34 + 54 + 64 ) 20024 k= 20024 k+^1 ,for allk≥1.
816.Ifx≤y≤z, then since 4x+ 4 y+ 4 zis a perfect square, it follows that the number
1 + 4 y−x+ 4 z−xis also a perfect square. Then there exist an odd integertand a positive
integermsuch that
1 + 4 y−x+ 4 z−x=( 1 + 2 mt)^2.It follows that
4 y−x( 1 + 4 z−x)= 2 m+^1 t( 1 + 2 m−^1 t);hencem= 2 y− 2 x−1. From 1+ 4 z−x=t+ 2 m−^1 t^2 , we obtain
t− 1 = 4 y−x−^1 ( 4 z−^2 y+x+^1 −t^2 )= 4 y−x−^1 ( 2 z−^2 y+x+^1 +t)( 2 z−^2 y+x+^1 −t).Since 2z−^2 y+x+^1 +t>t, this equality can hold only ift=1 andz= 2 y−x−1. The
solutions are of the form(x, y, 2 y−x− 1 )withx, ynonnegative integers.
817.With the substitutionu= 2 x+3,v= 2 y+3,w= 2 z+3, the equation reads
u^2 +v^2 +w^2 = 7.By eliminating the denominators, it is equivalent to show that the equation
U^2 +V^2 +W^2 = 7 T^2has no integer solution(U,V,W,T) = ( 0 , 0 , 0 , 0 ). Assuming the contrary, pick a
solution for which|U|+|V|+|W|+|T|is minimal. Reducing the equality modulo 4, we