# question involving modulus (congrency)

• Dec 26th 2009, 11:42 AM
cooltowns
question involving modulus (congrency)
Hi

I need help with the following question

I have attached my question.

I haven't got a clue on how to start the question.

thanks
• Dec 26th 2009, 11:47 AM
tonio
Quote:

Originally Posted by cooltowns
Hi

I need help with the following question

I have attached my question.

I haven't got a clue on how to start the question.

thanks

First, you prove that any integer to the 4th power is 0,1,2,or 4 mod 7, and then you reduce the given question modulo 7 and get that there's no solution to it and thus there's no integere solution for it, either.

Tonio
• Dec 26th 2009, 11:50 AM
cooltowns
can you please show me how to do it, i would appreichate this. I have a few questions which are similar to this. So if i have fully worked out solution to one i can try others on my own.

thanks
• Dec 26th 2009, 05:10 PM
HallsofIvy
You solve it by doing it! What is $\displaystyle 0^2 (mod 7)$? What is $\displaystyle 1^2 (mod 7)$? What is $\displaystyle 2^2 (mod 7)$? What is $\displaystyle 3^2 (mod 7)$? What is $\displaystyle 4^2 (mod 7)$? What is $\displaystyle 5^2 (mod 7)$? What is What is $\displaystyle 6^2 (mod 7)$?

Now reduce $\displaystyle 7x^5+ 3y^4= 2$ "(mod 7)". Since 7= 0 (mod 7) that reduces to $\displaystyle 3y^4= 3(y^2)(y^2)= 2$ (mod 7). Is there any integer y for which that is true? Replace $\displaystyle y^2$ with 0, 1, 2, and 4 to see.
• Dec 27th 2009, 12:39 AM
cooltowns
could you please explain how the $\displaystyle 3y^4=3(y^2)(y^2)=2(mod7)$

i don't understand where the 2 came from and why you have chosen mod7 in particular.

thanks
• Dec 27th 2009, 01:26 AM
CaptainBlack
Obviously:

$\displaystyle 7x^5+3y^4 \equiv 3y^4\ (\text{mod}\ 7)$

Now we know that $\displaystyle y^4\equiv 0, 1,2, \text{ or } 4\ (\text{mod} \ 7)$

So what are the possible values of $\displaystyle 3y^4\ (\text{mod}\ 7)$ and are any of these congruent to $\displaystyle 2 \ (\text{mod}\ 7)$

CB
• Dec 27th 2009, 01:56 AM
cooltowns
but why 2(mod7). As i understand the mod 7 comes from the equation but could you please explain where the 2 came from ?

thank you
• Dec 27th 2009, 02:26 AM
CaptainBlack
Quote:

Originally Posted by cooltowns
but why 2(mod7). As i understand the mod 7 comes from the equation but could you please explain where the 2 came from ?

thank you

It's the right hand side of your d**n problem:

$\displaystyle 7x^5+3y^4=2$

There is a bit of a hint also in the question that $\displaystyle \text{mod 7}$ is involved.

To prove that an equation has no integer solutions it is sufficient to show that it has no solution modulo some integer.

(As a hint in future read your own problem when you post it)

CB
• Dec 27th 2009, 04:19 AM
cooltowns
thanks for all the replies.

(Wink)