[Soumyadeep mandal]

when a four degree equation cuts the x axis at 3 points (provided that all the roots are real) it means one of these three roots have a multiplicity of 2. hence note that for the function :

f(x)=(x-a)^2(x-b)(x-c);       f(a)= 0 also: f"(a)=0 Implies maxima or minima.... hence proved.

 

 

[Soumyadeep mandal]

The answer is none of the integers as 1995,1999,2003 each are congruent to 3 mod 4 whereas the sum of two squares are always congruent to 0,1,2 mod 4. hence none of them can be expressed.

 

 

[Soumyadeep mandal]

Note that 2007 is an odd number hence,

x and y must be of opposite parity, let x be of the form 2m+1 and y be of the form 2n. (Opposite parity).

hence the LHS becomes :

(2m+1)^2 + (2n)^2   =.  4m^2 + 4m +1 + 4n^2   =  4k+1 (for some integer k)

Therefore LHS is congruent to 1 mod 4 ...... But note that 2007 is congruent to 3 mod 4 ....contradiction !

Hence there are no integral solutions for this equation...Please correct me if my solution is wrong.

Thank you

 

[Soumyadeep mandal]

thanks

 

[Soumyadeep mandal]

well even I am getting the answer as 2^19 + (1/2)*(20C10).... I think the sum is till (20C9) only, then only we can get option 4 as the answer ...

 

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