Polynomial Multiple of Five

First note that $y$ must be one of $0,1,2,3,4$ modulus $5$ .

If $y \equiv 0 \pmod{5}$ then $y^{2} + y + 1 \equiv 0 + 0 + 1 \equiv 1 \pmod{5}$ .

If $y \equiv 1 \pmod{5}$ then $y^{2} + y + 1 \equiv 1 + 1 + 1 \equiv 3 \pmod{5}$ .

If $y \equiv 2 \pmod{5}$ then $y^{2} + y + 1 \equiv 4 + 2 + 1 \equiv 2 \pmod{5}$ .

If $y \equiv 3 \pmod{5}$ then $y^{2} + y + 1 \equiv 4 + 3 + 1 \equiv 3 \pmod{5}$ .

If $y \equiv 4 \pmod{5}$ then $y^{2} + y + 1 \equiv 1 + 4 + 1 \equiv 1 \pmod{5}$ .

Since in none of these cases is the remainder $0$ , we can conclude that $y^{2} + y + 1$ is never divisible by $5$ .

(Note that since the remainder is never $4$ , we can also conclude that $y^{2} + y + 2$ is never divisible by $5$ either.)

$y^2+y+1$ is divisible by 5 when the units digit of $y^2+y$ or $y\cdot(y+1)$ is either 4 or 9. Try to multiply 2 successive numbers from 0 to 9 and you will never find 4 or 9 as the units digit.

$\displaystyle \Rightarrow y^2 + y + 1 \equiv 3,1,2,1,1 (mod 5)$ for $y =1,2,3,4,5$ respectively. This cycle repeats itself after every 5 numbers.

$\displaystyle \Rightarrow y^2 + y + 1 \equiv -4,-2, 2, 3, 1 (mod 5)$ for $y = -5,-4,-3,-2,-1$ respectively. Even this cycle repeats itself after every 5 numbers.

For $y=0$ , the expression equals 1, which is not divisible by 5.

Hence, $y^2 + y + 1$ is never divisible by 5.

We factor the expression as $y(y+1)+1$ . Since all numbers divisible by 5 end in either a 0 or 5, then $y(y+1)$ must end in 9 or 4.

The product of two consecutive integers is never odd, because we are always multiplying an even number with an odd number. So, $y(y+1)$ can never end in 9.

If we write out all the possibilities for the possible last digits of the product of two consecutive numbers, we see that the only possibilities are 0, 2, and 6. So, $y(y+1)$ cannot end in 4 either.

Since $y(y+1)$ cannot end in 9 or 4, $y(y+1)+1=y^2+y+1$ cannot end in 0 or 5. Thus, it will never be divisible by 5.

Suppose $y^2+y+1$ is not divisible by $5$ . Then $y\pm 5n \quad \forall n \in \mathbb{N}$ shares the same property. Thus, every $5^{\text{th}}$ integer from $y$ shares the same property. It then remains to verify that $5$ consecutive integers do not share this property, which is trivial, and we are done. Thus, there cannot exist an integer that satisfies this property.

Okay, so note that any integer can be written in the form $5k + n$ , where $n$ is any number from $0$ to $4$ .

Plugging in $y = 5k + n$ into $y^{2} + y + 1$ , we get: $(25k^{2} + 10kn + n^{2}) + (5k + n) + 1 = 5*(5k^{2} + 2kn + k) + (n^{2} + n + 1)$ . Thus, if $y = 5k + n$ yields a multpole of $5$ , then $n^{2} + n + 1$ must also be divisible by $5$ .

For $n = 0$ , we have: $0^{2} + 0 + 1 = 1$ , which is not divisible by 5.

For $n = 1$ , we have: $1^{2} + 1 + 1 = 3$ , which is also not divisible by $5$ .

Repeating the process for $2, 3$ and $4$ will yield, respectively, $7, 13$ and $21$ , neither of which are divisible by $5$ . Thus, no numer of the form $y^{2} + y + 1$ can be divided by $5$ .

(EDIT: Changed an incorrect word I used)

Y²+y+1 can't be expressed in form of 5m