[MAA_96]

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

[ghostwalker]

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?

[elldeegee]

Could you not use the quadratic formula?





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

[MAA_96]

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

[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:

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9

[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

[bananarama2]

(Original post by 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

Clever. I wasn't thinking.

[Blutooth]

(Original post by wcp100)
Clever. I wasn't thinking.

Thank you, but if you want clever I just solved an IMO problem in the summer maths thread. Took me 2 days to work out and I was thinking

[ghostwalker]

(Original post by Blutooth)
...

Nice; for some reason I was getting 2 when thinking of 7 (mod 3) ; brain dead.

Defo. not "secondary school".

+rep.