IIT JEE , AIEEE Forum - Mathematics , Algebra

feynmann

If possible let p( b ) = 0 for an integer b .

case - I ( b< 1 )

Now , by division algorithm , we may write

p ( x ) = ( x - b ) Q ( x ) ( Q ( x ) being a polynomial with integer coeff s ) ..............( 1 )

Now we have p ( 1 ) = ( 1 - b ) Q ( 1 )

p ( 2 ) = ( 2 - b ) Q ( 2 )

...............................................

p( c ) = ( c- b ) Q ( c )

Now see that all Q ( i ) s are integer ............ ( 2 )

& among ( 1 - b ) , ( 2- b ) , ............ ( c- b ) { They are c consecutive integers & all 0 or -ve ) at least one must be divisible by c ............ ( 3 )

So from ( 2 ) & ( 3 ) it follows that at least one of p( 1 ) , p( 2 ) ,............p( c )

must be divisible by c .

But according to the given condition this is simply wrong .

So our hypothesis is wrong ( Reductio ad absurdum !!!!!!! )

So p( b ) is not zero for any negative integer b . ( Proved.........1 )

case - II

If b >c we take the factor to be ( b - x ) & all of ( b- 1 ) ( b- 2 ) ,, ..... ( b-c ) are positve and consecutive integers so atleast one must be divisible by c .

So our hypothesis is wrong .

so p( b ) is not 0 for any b>c. ( proved ........ 2 )

case - III

Now we have to prove for 1 < b <c

we see that for this case there is a b < c for which p( b ) = 0 , so p( b ) is divisible by c which is again against the given condition .

again our hypothesis is wrong .

so p( b ) is non zero for 1 < b < c ( proved ......... 3 )

So we havecompleted the proof for all possible cases.

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