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Determine all positive integers

Determine all positive integers \((x,n)\) so that \(x^n+2^n+1\) is divisor of \(x^{n+1}+2^{n+1}+1\)

Note by Idham Muqoddas
4 years ago

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Interesting problem but very messy .

I will show that only solutions are \( (11,1),(4,1) \) . First if \( n=1 \) then \( x+3 \mid x^2+5 \implies x+3 \mid x^2−9+14 \implies x+3 \mid 14 \implies x=11,4 \) Thus \( (11,1) \) and \( (4,1) \) are two solutions. .

Now suppose \( x \ge 4 \)
We will prove that \[ (x-1)(x^n+2^n+1) < x^{n+1}+2^{n+1}+1 < x ( x^n+2^n+1) \] Which can be verified by seeing that , \[ x^n=\frac{x^n}{2}+ \frac{x^n}{2} \ge \frac{4^n}{2}+\frac{x^2}{2} \] And , \[ (x-1)(x^n+2^n+1) = x^{n+1}+x(2^n+1) - (x^n+2^n+1) < \\ x^{n+1}+\frac{x^2+(2^n+1)^2}{2} - (x^n+2^n+1) = \\ x^{n+1}+\frac{4^n+2 \cdot 2^n+1+x^2}{2} - (x^n+2^n+1) < \\ x^{n+1}+ 2^{n+1}+x^n+2^n+2 - (x^n+2^n+1) < x^{n+1}+2^{n+1}+1 \] And \[ x(x^n+2^n+1) > x^{n+1}+2^{n+1}+1 \] Is quite obvious. So there are no solutions , for the case when \( x\ge 4 \) Now suppose \( x \in 1,2,3 \) then ,

\[ \text{If} \ \ x=1 \implies 1^n+2^n+1<2+2^n+1<2+2^{n+1}+1<2(1+2^n+1) \] \[ \text{If} \ \ x=2 \implies 2^n+2^n+1<2^{n+1}+2^{n+1}+1<2(2^n+2^n+1) \] and \[ \text{If} \ \ x=3 \implies 2(3^n+2^n+1)<3^{n+1}+2^{n+1}+1<3(3^n+2^n+1) \] So for \( x=1,2,3 \) also there are no solutions.

So Only possible solutions are \( (11,1) , (4,1) \) . \( \Box \).

PS : Comment , if there is any mistake or if any-part is not explained well Shivang Jindal · 4 years ago

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@Shivang Jindal how do you get this? \[x\left(x^n+2^n+1\right)>x^{n+1}+2^{n+1}+1\] Idham Muqoddas · 4 years ago

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@Idham Muqoddas Clearly, \(x \cdot x^n = x^{n+1}\), and \(x \cdot 2^n > 2 \cdot 2^n = 2^{n+1}\), and \(x \cdot 1 = x > 1\).

Add these to get \(x(x^n+2^n+1) > x^{n+1}+2^{n+1}+1\). Jimmy Kariznov · 4 years ago

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