數學algebra一問

Solution: if y>5
y! = 0 (mod 9)
R.H.S.=3(mod 9)
L.H.S. =
= 0,1,4,9 or 16 (mod 9)
= 0, 1, 4, 0, 7 (mod 9)
Contradiction results
So therefore y ≦5
Substituting each of them into the equation
We get the only solution is (x,y) = (45, 4)


2007-01-01 12:42:35 補充：
Sorry變了亂碼，我再post過……Solution: if y 5y! = 0 (mod 9)R.H.S.=3(mod 9)L.H.S. = 0,1,4,9 or 16 (mod 9)= 0, 1, 4, 0, 7 (mod 9)Contradiction resultsSo therefore y ≦5Substituting each of them into the equationWe get the only solution is (x,y) = (45, 4)

Both KR9497 and 發霉pizza have given very excellent answers, what I am going to do is just a summary of their answers.

Base on the properties of y! , we will approach the problem by considering y first.
For y=1 and 2, it is clear that there is no solution.

For y >= 3, we have x^2 = 2001 +y! =3x23x29 +y! is divisible by 3.
This implies that 9 divides x^2

For y >5, 9 divides y! and hence 9 divides x^2 -y!
However, 2001= 222x9 +3
This implies that the range of solution of y is 3 =< y =< 5

Substituting y=3 , 4 and 5 respectively into the equation x^2-y!=2001 , we can conclude that (x,y) = (45, 4) is the only solution.

This means that there is only one pair of solution for the equation x^2-y!=2001 .

2007-01-01 14:54:21 補充：
p.s. 發霉pizza .... 睇你的功夫一點也不像中學生，好!

(45, 4)

program noname;
var x,y:integer;
function factorial(a:integer):real;
var i:integer;ans:real;
begin
ans:=1;
for i:= 1 to a do begin
ans := ans*i;
end;
factorial := ans
end;
begin
for x:= 1 to 100000 do
for y:= 1 to 40 do
if x*x - factorial(y) = 2001 then writeln(x:5,y:5);
readln
end.

The answer is found by this program