Year 12s: what will be the first way you research your future options? >>

proof for equation with no solution

How to prove that there is no solution for this equation in natural numbers:

Consider modulo arithmetic.

I could tell you what base works but not the justification for choosing it, as my number theory is too rusty.

Is this really "Secondary school" level?

(edited 11 years ago)

Reply 2

elldeegee

Could you not use the quadratic formula?

3n^2 + 3n + 7 = k^3[br]3n^2 + 3n + 7 - k^3 = 0

\dfrac{-b\pm \sqrt{b^2 - 4ac}}{2a}[br]= \dfrac{ -3 \pm \sqrt{9 - 12(7-k^3)}}{6} ...

I then wouldn't really know where to go from there, sorry.
I just remember seeing something like this a couple of ears ago.

Reply 3

MAA_96

But quadratic formula is not proving anything i think ,i know that we can prove it by modulo but how ?

Reply 4

ghostwalker

Original post by MAA_96

i know that we can prove it by modulo but how ?

With the right base, the LHS can only be equal to certain values, and the RHS can only be equal to certain other values; and those two sets of values do not intersect, and hence there is no solution.

Base is

Spoiler

Reply 5

Blutooth

3n^2+3n+7 =1 mod 3. Thus K^3 must be 1 mod 3. let K=3m+1 imples
3n^2+3n+7=27(m^3+m^2)+9m+1
n^2+n+2=9(m^3+m^2)+3m
rhs=0 mod 3. n^2+n+2 is either 2 or 1 mod 3, never 3 by considering residue classes
done

(edited 11 years ago)

Reply 6

bananarama2

Original post by Blutooth