2x+1 is a perfect square but the following x+1 integers are not.

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Sumgato

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Sumgato

#1 h Mar 17, 2018, 4:30 PM • 2 Y

Y by Adventure10, Mango247

Find all positive integers $x$ such that $2x+1$ is a perfect square but none of the integers $2x+2, 2x+3, \ldots, 3x+2$ are perfect squares.

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TuZo

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TuZo

#2 h Mar 17, 2018, 4:41 PM • 2 Y

Y by Adventure10, Mango247

We have $x=a^2$ , and $2x+1=b^2$ , so $b^2-2a^2=1$ , and this is a Pell equation, with infinite many solution.
On the other hand, we must have $m^2<2x+2$ , and $3x+2<(m+1)^2$ , so we get $\frac{{{m}^{2}}-2}{2}<{{a}^{2}}<\frac{{{(m+1)}^{2}}-2}{3}$ (*), so we have ${{m}^{2}}-4m-4<0$ , thus $m\in \left\{ 0,1,2,3,4 \right\}$ , and we must verify these numbers in (*), and we get the $a$ , and the $x$ .

This post has been edited 6 times. Last edited by TuZo, Mar 17, 2018, 4:53 PM

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nikolapavlovic

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nikolapavlovic

#3 h Mar 17, 2018, 4:49 PM • 2 Y

Y by Adventure10, Mango247

$2\sqrt{2x+1}+1\geq x+1$ which holds for $\mathcal{O}(1)$ numbers.

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Mr.Bash

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Mr.Bash

#4 h Mar 17, 2018, 4:49 PM • 8 Y

Y by Involution, othman, pro_4_ever, Wizard_32, AlastorMoody, Ya_pank, Adventure10, Mango247

Note that the next square number after $2x+1$ is $(\sqrt{2x+1} + 1)^2$ . As none of the consecutive positive integers $2x+2, 2x+3, \ldots, 3x+2$ are squares, we must have $(\sqrt{2x+1} + 1)^2 > 3x+2$ . This simplifies to $x^2-8x-4<0$ . Hence, $(x-4)^2 <20$ . As $x$ is natural, $\mid x-4\mid$ lies from $1$ to $4$ . So, $x$ can take the values $1,2,3,4,5,6,7,8$ . Out if these, only $4$ gives $2x+1$ to be a perfect square. Note that this solution for $x$ indeed satisfies the conditions.
$\blacksquare$

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othman

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othman

#5 h Mar 18, 2018, 2:08 AM • 2 Y

Y by Adventure10, Mango247

TuZo wrote:

We have $x=a^2$ , and $2x+1=b^2$ , so $b^2-2a^2=1$ , and this is a Pell equation, with infinite many solution.
On the other hand, we must have $m^2<2x+2$ , and $3x+2<(m+1)^2$ , so we get $\frac{{{m}^{2}}-2}{2}<{{a}^{2}}<\frac{{{(m+1)}^{2}}-2}{3}$ (*), so we have ${{m}^{2}}-4m-4<0$ , thus $m\in \left\{ 0,1,2,3,4 \right\}$ , and we must verify these numbers in (*), and we get the $a$ , and the $x$ .

according to the problem statement, x doesn't have to be a perfect square, right?

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Mahdi_Mashayekhi

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Mahdi_Mashayekhi

#6 h Feb 13, 2022, 8:36 AM

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Let $2x+1 = y^2$ so $y = \sqrt{2x+1}$ .
Note that whole problem says $(y+1)^2 - y^2 > (3x+2) - (2x+1) = x+1$ so $2y+1 > x+1$ so $2y > x$ .
$2\sqrt{2x+1} > x$ so $4(2x+1) > x^2$ so $4 > x(x-8)$ so $x < 9$ .
Now we can check that only for $x = 4$ , $2x+1$ is perfect square.

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MuradSafarli

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MuradSafarli

#7 h Mar 23, 2025, 3:04 PM

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It is clear that the number $2x+1$ is an odd number, and if it is a perfect square, then it must be in the form $(2n+1)^2$ for some natural number $n$ .

$2x+1 = (2n+1)^2$

This simplifies to $x = 2n^2 + 2n$ .

Also, none of the numbers from $2x+2$ to $3x+2$ are perfect squares. Thus, we have the inequality:

$3x+2 < (2n+2)^2$ .

Substituting $x = 2n^2 + 2n$ into the inequality, we get:

$n^2 < n+1$ .

The only natural number satisfying this inequality is $n = 1$ .

Therefore, $x = 4$ .

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Davut1102

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Davut1102

#8 h Mar 23, 2025, 4:03 PM

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Let $a^2=2x+1$ .Then $\frac{(a+1)^2}{a^2} \ge \frac{3x+2}{2x+1} \ge \frac{3}{2}$ and the rest is trivial.

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Mathdreams

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Mathdreams

#9 h Mar 23, 2025, 7:35 PM

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Solution

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lksb

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lksb

#10 h Mar 23, 2025, 9:01 PM

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Let $2x+1=n^2$ , then, if none of the numbers $2x+2, 2x+3,\dots, 3x+2$ are perfect squares, then $(n+1)^2>3x+2$
From that, we have $3x+2<(n+1)^2=n^2+2n+1=2x+2n+2$ $\implies \boxed{x<2n}$ Then, $n^2=2x+1<4n+1\iff \boxed{n\leq 4}$
Testing leads gives us $x=0$ and $x=4$ as solutions

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