Answer with Step-by-step explanation:Let Z be a set of integers We have to prove that [tex]x^2-5x-1[/tex] is odd for all value of x [tex]\in[/tex]Z.There are two cases 1.When x is an odd integer number 2.when x is even integer number 1.When x is an odd integer number When x is positive odd integer Then [tex]x^2[/tex] is positive odd number (because square of odd positive number is always positive an odd number )5x is also odd number -5x -1 = -even number Suppose x=3Then -15-1=-16Odd number - even number=Odd number Hence ,[tex]x^2-5x-1[/tex] is an odd number .When x is negative odd number Then x square is positive odd number and 5x is negative odd term termTherefore,odd number + odd number -1=Even number -1=Odd number Hence [tex]x^2-5x-1 [/tex] is an odd number.2.When x is an even number When x is positive even number x square is positive even number and 5x is positive even number Even number -Even number -1 =Even number -odd number =Odd number Suppose x=4 [tex]x^2-5x-1[/tex]=16-20-1=-5=Odd number When x is negative even numberThen x square is positive even number and 5x is negative even number [tex] x^2-5x-1[/tex]=Even number +Even number -1=Even number -1=Odd number Hence, for all elements of Z[tex]x^2-5x-1[/tex]is odd.