It Is Being Given That (5^32+1) Is Completely Divisible By A Whole Nu

1.	It Is Being Given That (5^32+1) Is Completely Divisible By A Whole Number. Which Of The Following Numbers Is Completely Divisible By This Number?
Answer» LET 5^32=x. Then (5^32+1)=(x+1). Let (x+1) be COMPLETELY divisible by the WHOLE NUMBER Y. then (5^96+1)=[(5^32)^3+1]=>(x^3+1)=(x+1)(x^2x+1) which is completely divisible by Y. since (x+1) is divisible by Y. Let 5^32=x. Then (5^32+1)=(x+1). Let (x+1) be completely divisible by the whole number Y. then (5^96+1)=[(5^32)^3+1]=>(x^3+1)=(x+1)(x^2x+1) which is completely divisible by Y. since (x+1) is divisible by Y.

It Is Being Given That (5^32+1) Is Completely Divisible By A Whole Number. Which Of The Following Numbers Is Completely Divisible By This Number?

Answer»

LET 5^32=x.

Then (5^32+1)=(x+1). Let (x+1) be COMPLETELY divisible by the WHOLE NUMBER Y.

then (5^96+1)=[(5^32)^3+1]=>(x^3+1)=(x+1)(x^2x+1) which is completely divisible by Y.

since (x+1) is divisible by Y.

Let 5^32=x.

Then (5^32+1)=(x+1). Let (x+1) be completely divisible by the whole number Y.

then (5^96+1)=[(5^32)^3+1]=>(x^3+1)=(x+1)(x^2x+1) which is completely divisible by Y.

since (x+1) is divisible by Y.