Determine all positive integers

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

#NumberTheory #MathProblem #Math

Note by Idham Muqoddas
7 years, 10 months ago

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4 votes

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Comments

Interesting problem but very messy .

I will show that only solutions are (11,1),(4,1) (11,1),(4,1) . First if n=1 n=1 then x+3∣x2+5  ⟹  x+3∣x2−9+14  ⟹  x+3∣14  ⟹  x=11,4 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) (11,1) and (4,1) (4,1) are two solutions. .

Now suppose x≥4 x \ge 4
We will prove that (x−1)(xn+2n+1)<xn+1+2n+1+1<x(xn+2n+1) (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 , xn=xn2+xn2≥4n2+x22 x^n=\frac{x^n}{2}+ \frac{x^n}{2} \ge \frac{4^n}{2}+\frac{x^2}{2} And , (x−1)(xn+2n+1)=xn+1+x(2n+1)−(xn+2n+1)<xn+1+x2+(2n+1)22−(xn+2n+1)=xn+1+4n+2⋅2n+1+x22−(xn+2n+1)<xn+1+2n+1+xn+2n+2−(xn+2n+1)<xn+1+2n+1+1 (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(xn+2n+1)>xn+1+2n+1+1 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≥4 x\ge 4 Now suppose x∈1,2,3 x \in 1,2,3 then ,

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

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

PS : Comment , if there is any mistake or if any-part is not explained well

Shivang Jindal - 7 years, 10 months ago

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how do you get this? x(xn+2n+1)>xn+1+2n+1+1x\left(x^n+2^n+1\right)>x^{n+1}+2^{n+1}+1

idham muqoddas - 7 years, 10 months ago

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Clearly, x⋅xn=xn+1x \cdot x^n = x^{n+1}, and x⋅2n>2⋅2n=2n+1x \cdot 2^n > 2 \cdot 2^n = 2^{n+1}, and x⋅1=x>1x \cdot 1 = x > 1.

Add these to get x(xn+2n+1)>xn+1+2n+1+1x(x^n+2^n+1) > x^{n+1}+2^{n+1}+1.

Jimmy Kariznov - 7 years, 10 months ago
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