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

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?

• Do you want X to be an integer as well? If not, there are infinite solutions. – Peter Foreman Jan 29 '19 at 16:47

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 $$x\equiv 3 \pmod{5}$$.

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?

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.