Finding Integers for x^2-2x-3 Divisibility by 8: Help with Mod Problem

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SUMMARY

The integers for which the expression x² - 2x - 3 is divisible by 8 are those that satisfy the conditions x ≡ 3 (mod 8) or x ≡ 7 (mod 8). The solution involves factoring the expression into x(x - 2) and determining when this product equals 3 modulo 8. Exhaustive checking of integer values from 0 to 7 reveals that only x = 3 and x = 7 yield the desired result, leading to the general solutions of the form x = 3 + 8n or x = 7 + 8n, where n is any integer.

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Homework Statement



For what integers is x^2-2x-3 divisible by 8.

give answer x = a list of natural numbers (mod 8)

Homework Equations





The Attempt at a Solution



I thought x(x-2) = 3 (mod 8) and tried to find all values that satisfied this. But got really stuck.
I also tried to find values that gave a reminder of zero when the quadratic was divided by 8, which i think was 7, 11, 15 but this was not correct.
I thought this was quite difficult as i thought my solutions made sense, but i must have misunderstood it somewhere. Thanks! its much appreciated.
 
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Factorize the left hand side first: it is x(x-2). The product has to be 3(mod 8), so the factors have to be either {1(mod 8) * 3(mod 8)} or {(-1)(mod 8) * (-3)(mod 8)}. Which of x and x-2 can be 1 (mod 8) so the other is 3 (mod 8)? The same with -1 and -3. ehild
 
Last edited:
Since this is "mod 8", it's not too difficult to "exhaust" the cases:
if x= 0 then 0(0- 2)= 0 (mod 8), not 3
if x= 1 then 1(1- 2)= -1= 7 (mod 8), not 3
if x= 2 then 2(2- 2)= 0 (mod 8), not 3
if x= 3 then 3(3- 2)= 3 (mod 8) !

if x= 4 then 4(4- 2)= 8= 0 (mod 8), not 3
if x= 5 then 5(5- 2)= 15= 7 (mod 8), not 3
if x= 6 then 6(6- 2)= 24= 0 (mod 8), not 3
if x= 7 then 7(7- 2)= 35= 3 (mod 8)!

So x= 3 (mod 8) or x= 7 (mod 8) so any x of the form x= 3+ 8n or x= 7+ 8n, where 8 is any integer, satisfies this equation.
 

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