find the smallest positive integer x such that where x and y are positive integers
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1 0 0 0 0 0 1 = 1 0 6 + 1 = ( 1 0 2 + 1 ) ( 1 0 4 − 1 0 2 + 1 ) = 1 0 1 × 9 9 0 1
Provided that both 101 and 9901 prime number and x + y > x − y , there will be case where ( x + y , x − y ) = ( 9 9 0 1 , 1 0 1 ) , ( 1 0 0 0 0 0 1 , 1 ) . To find the minimum x , we will choose the lowest value of ( x + y ) + ( x − y ) , which is ( x + y , x − y ) = ( 9 9 0 1 , 1 0 1 ) . Solve then x = 5 0 0 1