Another 800 problem

[lime777]

Is there a value of x such that x^2 + x + 1 is a multiple of 5? Explain.

Answers will be posted later.

[hemant_rajput]

Is there a value of x such that x^2 + x + 1 is a multiple of 5? Explain.

Answers will be posted later.

I don't think this is a 800 level question. It hardly took me 10 sec to solve.

here is my solution

eq is (x^2 + x + 1)/5 = n, where n is an integer

any no. which is not a multiple of 5 will leave remainder 1 or 2 or 3 or 4.

for any no. x which leave remainder 1 when divided by 5, when x^2 + x + 1 divided by 5 we will get a remainder of 1*1 + 1 + 1 = 3. so x is not x^2 + x + 1 is not divisible for any value of x which will give remainder 1.

now,
any no. x which leave remainder 2 when divided by 5, when x^2 + x + 1 divided by 5 we will get a remainder of 2*2 + 2 + 1 = 2. so x is not x^2 + x + 1 is not divisible for any value of x which will give remainder 1.

similarly,

any no. x which leave remainder 3 and 4 when divided by 5, when x^2 + x + 1 divided by 5 we will get a remainder of 3*3 + 3 + 1 = 3 and 4*4 + 4 + 1 = 1. so x is not x^2 + x + 1 is not divisible for any value of x which will give remainder 3 and 4.

now for any value of x which is divisible by 5 we will always get remainder 1.


hence there is no value of x for which x^2 + x + 1 is a multiple of 5

[Anurag@Gurome]

Is there a value of x such that x^2 + x + 1 is a multiple of 5?

If x is not an integer, (x^2 + x + 1) also will not be integer.

Now, if x is an integer, (x^2 + x + 1) = x(x + 1) + 1 = Product of two consecutive integers + 1
As product of two consecutive integers is always even, (x^2 + x + 1) will be odd. Hence, if (x^2 + x + 1) is a multiple of 5 it must be an odd multiple of 5, i.e. the unit's digit of (x^2 + x + 1) must be 5.

Hence, the unit's digit of x(x + 1) must be 4.
But there are no such consecutive integers such that their product ends with 4.

Hence, there is no x for which (x^2 + x + 1) is a multiple of 5.