Solve the congruence d x d + 2 = 14 8

The equation can be written as:

$$d^2 + 2 \equiv 14 \pmod{8}$$

$$d^2 \equiv 14 - 2 \pmod{8}$$ $$d^2 \equiv 12 \pmod{8}$$

To find $12 \pmod{8}$, we divide 12 by 8 and find the remainder: $12 = 1 \times 8 + 4$.
So, $12 \equiv 4 \pmod{8}$.
The congruence becomes:

$$d^2 \equiv 4 \pmod{8}$$

We need to find integers $d$ such that $d^2$ leaves a remainder of 4 when divided by 8. We can test values for $d$ from 0 to 7:

• If $d=0$, $d^2 = 0^2 = 0$. $0 \not\equiv 4 \pmod{8}$.
• If $d=1$, $d^2 = 1^2 = 1$. $1 \not\equiv 4 \pmod{8}$.
• If $d=2$, $d^2 = 2^2 = 4$. $4 \equiv 4 \pmod{8}$. This is a solution.
• If $d=3$, $d^2 = 3^2 = 9$. $9 \equiv 1 \pmod{8}$. $1 \not\equiv 4 \pmod{8}$.
• If $d=4$, $d^2 = 4^2 = 16$. $16 \equiv 0 \pmod{8}$. $0 \not\equiv 4 \pmod{8}$.
• If $d=5$, $d^2 = 5^2 = 25$. $25 \equiv 1 \pmod{8}$. $1 \not\equiv 4 \pmod{8}$.
• If $d=6$, $d^2 = 6^2 = 36$. $36 \equiv 4 \pmod{8}$. This is a solution.
• If $d=7$, $d^2 = 7^2 = 49$. $49 \equiv 1 \pmod{8}$. $1 \not\equiv 4 \pmod{8}$.

The values of $d$ that satisfy the congruence are $d \equiv 2 \pmod{8}$ and $d \equiv 6 \pmod{8}$.

The final answer is $d \equiv 2 \pmod{8} \text{ or } d \equiv 6 \pmod{8}$.