Number Theory 715800.Note that for any integerk, we can dissect thed-dimensional cube intokdpieces. If
we do this for two integersaandb, then performing the appropriate dissections we can
obtain(ad− 1 )x+(bd− 1 )y+1 cubes.
By Sylvester’s theorem for coprime positive numbersαandβ, the equationαx+βy=
nhas nonnegative solutions provided thatnis sufficiently large.
To complete the solution, we just have to findaandbsuch thatad−1 andbd−1 are
coprime. We can choose anyaand then letb=ad−1. Indeed,(ad− 1 )d−1 differs
from a power ofad−1 by 1, so the two numbers cannot have a common divisor.
801.There exist integersuandvsuch that the two sides in question area=u^2 −v^2 and
b= 2 uv. We are also told thata+b=k^2 , for some integerk. Then
a^3 +b^3 =(a+b)(a^2 −ab+b^2 )=k^2 ((u^2 −v^2 )^2 − 2 uv(u^2 −v^2 )+ 4 u^2 v^2 )
=k^2 (u^4 +v^4 − 2 u^3 v+ 2 uv^3 + 2 u^2 v^2 )=[k(u^2 −uv)]^2 +[k(v^2 +uv)]^2 ,and the problem is solved.
802.We guess immediately thatx =2,y =4, andz =2 is a solution because of
the trigonometric triple 3, 4 ,5. This gives us a hint as to how to approach the problem.
Checking parity, we see thatyhas to be even. A reduction modulo 4 shows thatxmust
be even, while a reduction modulo 3 shows thatzmust be even. Lettingx= 2 mand
z= 2 n, we obtain a Pythagorean equation
( 3 m)^2 +y^2 =( 5 n)^2.Becauseyis even, in the usual parametrization of the solution we should have 3m=
u^2 −v^2 and 5n=u^2 +v^2. From(u−v)(u+v)= 3 mwe find thatu−vandu+vare
powers of 3. Unlessu−vis 1,u=(u−v+u+v)/2 andv=(u+v−u+v)/2 are
both divisible by 3, which cannot happen becauseu^2 +v^2 is a power of 5. Sou−v=1,
u+v= 3 m, andu^2 +v^2 = 5 n. Eliminating the parametersuandv, we obtain the simpler
equation
2 · 5 n= 9 m+ 1.First, note thatn=1 yields the solution mentioned in the beginning. Ifn>1, then
looking at the equation modulo 25, we see thatmhas to be an odd multiple of 5, say
m= 5 ( 2 k+ 1 ). But then
2 · 5 n=( 95 )^2 k+^1 + 1 =( 95 + 1 )(( 95 )^2 k−( 95 )^2 k−^1 +···+ 1 ),which implies that 2· 5 nis a multiple of 9^5 + 1 = 2 · 52 ·1181. This is of course impossible;
hence the equation does not have other solutions.
(I. Cucurezeanu)