Advanced book on Mathematics Olympiad

722 Number Theory

find that|U|,|V|,|W|,|T|is even, hence(U 2 ,V 2 ,W 2 ,T 2 )is also a solution, contradicting

minimality. Hence the equation does not have solutions.

(Bulgarian Mathematical Olympiad, 1997)

818.Clearly,y=0 does not yield a solution, whilex=y=1 is a solution. We show

that there are no solutions withy≥2. Since in this case 7xmust give residue 4 when

taken modulo 9, it follows thatx ≡ 2 (mod 4). In particular, we can writex = 2 n,

so that

3 y= 72 n− 4 =( 7 n+ 2 )( 7 n− 2 ).

Both factors on the right must be powers of 3, but no two powers of 3 differ by 4. Hence

there are no solutions other thanx=y=1.

(Indian Mathematical Olympiad, 1995)

819.First solution: One can see immediately thatx=1 is a solution. Assume that there

exists a solutionx>1. Thenx!is even, so 3x!has residue 1 modulo 4. This implies that

the last digit of the number 2^3

x!

is 2, so the last digit of 3^2

x!

= 23

x!

+1 is 3. But this is

impossible because the last digit of an even power of 3 is either 1 or 9. Hencex=1is

the only solution.

Second solution: We will prove by induction the inequality

32

x!

< 23

x!

,

forx≥2. The base casex=2 runs as follows: 3^2
2
= 34 = 81 < 512 = 29 = 23
2
.
Assume now that 3^2
x!
< 23
x!
and let us show that 3^2
(x+ 1 )!
< 23
(x+ 1 )!
.
Raising the inequality 3^2
x!
< 23
x!
to the power 2x!·x, we obtain
(
32

x!)^2 x!·x

<

(

23

x!)^2 x!·x

<

(

23

x!)^3 x!·x

.

Therefore, 3^2

(x+ 1 )!

< 23

(x+ 1 )!

, and the inequality is proved. The inequality we just proved

shows that there are no solutions withx≥2. We are done.

Remark.The proof by induction can be avoided if we perform some computations. In-

deed, the inequality can be reduced to

32

x!

< 23

x!

and then to

x!<

log log 3−log log 2

log 3−log 2

= 1. 13588 ....

(Romanian Mathematical Olympiad, 1985)