From the constant term you’ll get an equation involving $m$ and $l$. Use it to decide whether $m$ and $l$ must be even or odd.

List the integer factor pairs of 8 (including sign). Think about whether $a$ and $c$ are odd/even in each case.

Using $b=al-mc$ and the parity information, decide whether $b$ can ever be divisible by 4.

In Desmos, enter lists for possible values:

A=[-8,-4,-2,-1,1,2,4,8]

Then compute the matching list:

C=8/A

(These are all integers because each value in A divides 8.)

Enter the divisors of 15:

L=[-15,-5,-3,-1,1,3,5,15]

Then compute:

M=-15/L

Create sliders for indices i and j (integers) so you can pick one value from each list:

• set i from 1 to length(A)
• set j from 1 to length(L)

Define:

a=A[i], c=C[i], l=L[j], m=M[j]

Then compute:

b=a*l-m*c

Move the sliders through different valid combinations and observe whether $b$ is divisible by 4. You can check by computing $b/4$: if the result is not an integer, then $b$ is not a multiple of 4.

Expand:

Match with $8x^2+bx+15$ to get

• $ac=8$
• $-ml=15\ \Rightarrow\ ml=-15$
• $b=al-mc$

Since $ml=-15$ is odd, both $m$ and $l$ must be odd integers.

Because $ac=8$, the integer pairs $(a,c)$ must make one of these parity situations happen:

• One of $a,c$ is odd and the other is even (e.g., $1\cdot 8$ or $8\cdot 1$, with possible negative signs).
• Both $a$ and $c$ are even (e.g., $2\cdot 4$ or $4\cdot 2$, with possible negative signs).

Recall $b=al-mc$ with $m,l$ odd.

Case 1: one of $a,c$ is odd and the other is even. Then $al$ is odd$\times$odd $=$ odd, while $mc$ is odd$\times$even $=$ even, so $b=$ odd $-$ even $=$ odd. Therefore, $b$ cannot be divisible by 4.

Case 2: both $a$ and $c$ are even. The only absolute factor pairs are $(2,4)$ or $(4,2)$.

• If $(a,c)=(2,4)$, then $b=2l-4m=2(l-2m)$. Here $l$ is odd and $2m$ is even, so $l-2m$ is odd. Thus $b$ is twice an odd number, so it is not divisible by 4.
• If $(a,c)=(4,2)$, then $b=4l-2m=2(2l-m)$. Here $2l$ is even and $m$ is odd, so $2l-m$ is odd. Again, $b$ is twice an odd number, so it is not divisible by 4.

In Case 1, $b$ is odd, so it is not divisible by 4. In Case 2, $b$ is twice an odd number, so it is also not divisible by 4.

Therefore, the statement that must be true is $b$ is not a multiple of 4.