INMO 2024 Problem 3

[Dr. Ashani Dasgupta]

Let \(p\) be an odd prime number and \(a, b, c\) be integers so that the integers
\[
a^{2023}+b^{2023}, \quad b^{2024}+c^{2024}, \quad c^{2025}+a^{2025}
\]
are all divisible by \(p\). Prove that \(p\) divides each of \(a, b\), and \(c\).

[Aritra]

Lets change the question a bit , suppose I ask you to proove the following ,

Given, \(a^{2023}+b^{2023}=0\),\(b^{2024}+c^{2024}=0\),\(a^{2025}+c^{2025}=0\) . Prove that , \(a\) ,\(b\) ,\(c\) all are equal to zero.

wahhh ! What a easy problem . Let work it out first !
If one of the ,\(a\) ,\(b\) ,\(c\) is zero then done .
so assume none of them are zero , then you can write ,\((\frac{a}{b})^{2023}=-1\), \((\frac{b}{c})^{2024}=-1\), \((\frac{a}{c})^{2025}=-1\).
On multiplication we get ,\(a^2=-bc\).
\(a^2=-bc\) \(\implies \) \(a^{4048}=b^{2024}c^{2024}\), on simplifying this implies\(b\) =\(c\) .

so the \(b^{2024}+c^{2024}=0\) truns into \(2b^{2024}=0 \), and we can conclude my claim.

Reading till can feel what we are going to do . yes! we will boil down the problem in modulo , modulo \(p\) . Think what does the

\(a^{2023}+b^{2023}\),\(b^{2024}+c^{2024}\),\(a^{2025}+c^{2025}\) means under modulo p . exactly they are nothing but each of them are zero in modulo p ! exactly the first conditon of our dummy problem . wonderful !! next we need to prove \(a\) ,\(b\) ,\(c\)  are divisible by p . what that mean in modulo \(p\) ! uff you are really clever ! exactly exactly , that ,means \(a\) ,\(b\) ,\(c\) all are equal to zero.That means we reached at out dummy problem .

then for what you are waiting for . write down , the entire solution taking modulo \(p\) !

but wait !! how the fraction make sense!! ohh I see , basically they are inverse modulo\( p\) . Because we know every nonzero element is invertible under modulo p .

now I completely copy the above prove taking modulo \(p\).

here I shall use the notaion \(Z_{p}\) to mean modulo \(p\).

so if in some place if I say \(n=0\) in \(Z_{p}\) . that will mean \(n\equiv 0\) modulo \(p\) , if I say \(n=1\) in \(Z_{p}\) . that will mean \(n\equiv 1\) modulo \(p\).  If I say \(\frac{m}{n}\) in \(Z_{p}\). I shall mean \(m.n^{-1}\) under modulo p e.t.c.

now the actual solution

If one of the \(a\) ,\(b\) ,\(c\) is divisible by \(p\) then we are done .
otherwise , we are assuming that none of them are divisible by \(p\).  So one thing is very clear that all the three numbers are invertible under multiplication \(Z_{p}\) .

Now the idea is simple , we shall bring the entire problem just in the residue class of \(p\) .

From now all the equality the following equality is in \(Z_{p}\).

in\(Z_{p}\), the given condition means , \(a^{2023}+b^{2023}=0\),\(b^{2024}+c^{2024}=0\),\(a^{2025}+c^{2025}=0\).

and the assumption that \(a\) ,\(b\) ,\(c\) are not divisible by p means they are non zero in \(Z_{p}\).

so we can say , \((\frac{a}{b})^{2023}=-1\), \((\frac{b}{c})^{2024}=-1\), \((\frac{a}{c})^{2025}=-1\).
On multiplication we get ,\(a^2=-bc\).

given that \(a^2=-bc\) implies , \(a^{4048}=b^{2024}c^{2024}\), But this implies b=c . (Fill up the details.)

but then ,\(b^{2024}+c^{2024}=0\) truns into \(2b^{2024}=0 \) , b is not equal to zero (beacuse we have assume that b is not divisble by p, hope you have not forgot the line \(n=0\) in \(Z_{p}\) . that will mean \(n\equiv 0\) modulo \(p\), now "something" modulo zero means what ? eaxtly ! divisible , but \(b\) is not divisible by \(p\), that is our primary assumption) . so , 2=0 in \(Z_{p}\) ,That is 2 is divisible by p  . Which is not possible for odd prime.

• This reply was modified 1 year, 5 months ago by Aritra.
• This reply was modified 1 year, 5 months ago by Aritra.
• This reply was modified 1 year, 5 months ago by Aritra.
• This reply was modified 1 year, 5 months ago by Aritra.
• This reply was modified 1 year, 5 months ago by Aritra.
• This reply was modified 1 year, 5 months ago by Aritra.
• This reply was modified 1 year, 5 months ago by Aritra.
• This reply was modified 1 year, 5 months ago by Aritra.

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