ISI MATHS QN

[tania]

 CONSIDER- X^2+Y^2=2007. HOW MANY SOLUTIONS (X,Y) EXIST SUCH THAT X AND Y ARE POSITIVE INTEGERS?

1) NONE
2) EXACTLY TWO
3) MORE THAN TWO BUT FINITELY MANY
4) INFINITELY MANY

THANKS

[onionknight]

The only thing I can think of is to eliminate some cases. if you take y2 to the other side you get
x2= 2007 - y2
Now lhs is a perfect square which means RHS should be one too. When you square a number the units digit of the square can only be 0(if the number being squared has 0at units place), 1 (if the number being squared has 1 or 9 at units place),4( if the number being squared has 2 or 8 at units place), 5 (if the number being squared has 5 at units place) 6(if the number being squared has 4 or 6 at units place) 9 (if the number being squared has 3 or 7 at units place).

Now when you subtract y2 from 2007 you can get 7,6,3,2,1,8 at units place. You can rule out cases where you get 7,3,2 and 8 which would be when y has 0,2,8,5,3,7. Now you get 4 cases:
x= 10a+1 Y=10b+6
x=10a+1 y=10b+4
x=10a+9 y=10b+6
x=10a+9 y=10b+6
if you plug these in you'll find some error
like if it's 10a+1 and 10b+4 RHS will be
100 a2+100b2+80b+20a which is divisible by 20 and rhs is 2007-16-1=1990 this is not divisible so RHS and LHS can't be equal. Similarly, for other cases. Or you can just check the 20 cases :P

[Rajat]

Answer should be none.
This is how I solved it.

write x2 + y2 as (x+y)2 - 2xy
factorize this as (x+y - sqrt(2xy))(x+y + sqrt(2xy))
2007 can be factorized as 9*223 or 3*669 or 1*2007  ...... (give it a general form A*B)

Equating the two sides, we get simultaneous equations as:

x + y - sqrt(2xy) = A
x + y + sqrt(2xy) = B

=> sqrt(2xy) = (B - A)/2   => xy = (B-A)^2/8
if x,y are to be integers, (B-A)^2 should be multiple of 8.
We find out that for any combination of A and B, this condition is never satisfied.
Hence there is no solution

[onionknight]

Square root of 2xy will not necessarily be an integer therefore the two terms in the product will not necessarily be integers either

[Rajat]

we are starting with the assumption that x and y are integers and then proving that they cannot be integers. That is the approach

[onionknight]

square root of 2xy might not be an integer even if x and y are integers, so the simplistic assumption you make while factorizing 2007 into a product of two integers and saying that the two terms on the LHS can be either of these is incorrect. There could be an infinite number of cases since the two terms on the LHS might not be integers even when x and y are.

[Jim_Moriarity]

None