Solving f(x)=3x^2+7x=5 for m=23 and m=25 using Completing the Square

[clueles]

please help me with this!

f(x)=3x^2+7x=5. find the solution of f(x)==0(mod m) for

m=23
m=25

the only thing i have done so far is completed the square

 

[robert Ihnot]

Completing the square gives (X+7/6)^2=109/36.

Bringing the 36 to the other side gives: (6X+7)^2 =109.

For modulo 23, you should consider quadratic reciprocity. For M=25, the problem is easier.

 

[CRGreathouse]

f(x)=3x^2+7x=5. find the solution of f(x)==0(mod m) for

m=23
m=25

the only thing i have done so far is completed the square

I'll assume you mean $$f(x)=3x^2+7x-5$$ (that is, you're not setting it to a boolean value).

I'm not convinced that completing the square is relevant here. A quick use of quadratic reciprocity (don't worry, skip it if you haven't seen it) shows that there are no solutions modulo 23:

$$(6x+7)^2\equiv109\equiv17\pmod{23}$$

$$\left(\frac{17}{23}\right)=\left(\frac{23}{17}\right)=\left(\frac{6}{17}\right)=\left(\frac{2}{17}\right)\left(\frac{3}{17}\right)=\left(\frac{17}{3}\right)=\left(\frac{2}{3}\right)=-1$$

This doesn't mean anything, though, just that the real solution isn't on one of the modular solutions.

Look at it this way: replacing $$x$$ with $$x+1$$ you have $$f(x)=3x^2+6x+1+7x+1-5=(3x^2+7x-5)+(6x+2)$$. By choosing $$x$$ you should be able to cycle through all the congruence classes, since neither 23 nor 25 is divisible by 2 or 6.