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For the equation $y = 4x^2 + 8x + 5$, what are the values of x such that y/5 is an integer?

For example, if x = 3, $y = 4(3^2) + 8(3) + 5$ = 65

if x = 5, y = 145

if x = 8, y = 325

Is there a formula to determine values of x that will result in a value of y that is divisible by 5?

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    $\begingroup$ Do you want X to be an integer as well? If not, there are infinite solutions. $\endgroup$ – Peter Foreman Jan 29 '19 at 16:47
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You are looking for those integers $x$ such that $\frac{4x^2+8x+5}{5}$ is integer, that are the solutions of $$ x^2+2x \equiv 0 \pmod{5}. $$ Since $\mathbb{Z}_5$ is a field, this is equivalent to $x\equiv 0\pmod{5}$ or $x\equiv 3 \pmod{5}$.

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Just try $x=0,1,2,3,4$. Adding $5$ to $x$ will not change the remainder when $y$ is divided by $5$. You have shown that all $x$ that are equivalent to $0$ or $3 \pmod 5$ work. How about the others?

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I assume that you want X to also be an integer number.

Then you have $y = 4x^2 + 8x +5 = x(4x+8) + 5$

if you want y to be a multiple of 5, then $x(4x+8)$ must also be a multiple of 5. The values of x that will work are those of the form:

$$x = 5k$$ $$x = 5k + 3$$

where k is an integer. Since for $x = 5k+3$, $4x+8$ is congruent to 0 mod 5: $$4(5k+3)+8 \equiv 12+8 \equiv 0 \mod{5} $$

You can prove that if you take a number which is not congruent to 0 or 3 mod 5 then y won't be congruent to 0 mod 5 by working the expression $x(4x+8)$

Hope it was useful.

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