When is $32x+32$ a square number?

Question

I am trying to find out for what values for $x$ does the function $f(x)=32x+32$ return a square number? I found that this is the case for at least: $x \in \{64,1600,53824,1827904,62094400,2109381184\}$ but I can't see a pattern in those numbers.

Is there a formula which generates these values for $x$ instead of simply trying a lot of values?

I'm not sure about a formula, but didn't you miss one? If $x = 31$, then $32(31) + 32 = (32)(32)$ which is a square. — layman, Jun 15 '15 at 18:48
@randomgirl deleted her answer, but I thought it was useful. Basically she said that since $32 = 2^{5}$, then $32x + 32 = 2^{5}(x + 1)$. So if $x + 1 = 2^{2k + 1}$ for any integer $k \geq 0$, then $2^{5}(x + 1)$ will be a perfect square. But actually $x + 1$ could equal infinitely many possibilities, since we could have $x + 1 = 2a^{2}$ for any integer $a$, or $x + 1 = 2a^{2k}b^{2j}c^{2n}$ for any integers $a,b,c$ and integers $k, j, n \geq 0$ (and so on). This should at least convince you that there are infinitely many solutions. — layman, Jun 15 '15 at 19:10

wlad · Accepted Answer · 2018-02-04T13:21:14.243

17

$$32(x + 1) = y^2 \implies 32 \mid y^2 \implies 4 \mid y$$ so let $y = 4z$. $$2(x+1) = z^2 \implies 2 \mid z$$ so let $z = 2w$ $$x + 1 = 2w^2 $$ so given any $w$ we can solve for $x$.

So the set of solutions is $$\{(2w^2 - 1, 8w) \mid w \in \Bbb Z\}$$

edited Feb 04 '18 at 13:21

answered Jun 15 '15 at 18:49

wlad

8,185

@user46944 Please see my last edit. – wlad Jun 15 '15 at 18:54
You need to exclude $w=0$, don't you? – Clement C. Jun 15 '15 at 18:56
1

@ClementC. If you want. I'm listing all solutions over the integers. – wlad Jun 15 '15 at 18:57
Oops, my bad. Sorry about that. – Clement C. Jun 15 '15 at 18:59
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In fact, $32|y^2$ already implies $8|y$ the two consequtive steps aren't necessary. – Redundant Aunt Jun 15 '15 at 18:59
what does the $8w$ in the last line mean? And I assume that $4|y$ means that $y$ is divisible by $4$? – Héctor van den Boorn Jun 15 '15 at 19:02
@HéctorvandenBoorn $4 \mid y$ means $4$ divides $y$, so the other way round. – wlad Jun 15 '15 at 19:02
@user109899 Thanks! Edited. – wlad Jun 15 '15 at 19:02
Excellent answer! thanks – Héctor van den Boorn Jun 15 '15 at 19:09
1

@user3491648 that's not the other way around, "y is divisible by 4" and "4 divides y" mean the same thing. – Jonathan Hebert Jun 27 '15 at 06:37
Perhaps I'm an Amateur, but it's not that obvious, at least to me, that $(32 \mid y^2) \implies (8 \mid y)$, versus, $(32 \mid y^2) \implies (4 \mid y)$, or even, $(32 \mid y^2) \implies (2 \mid y)$. Yet I see that – MaximusFastidiousIrreverence Sep 05 '17 at 21:32
Yet I see that if, $u \cdot 2^5=y^2$ then $$u=2^{2n+1} \cdot 3^{\alpha_2} \cdot 5^{\alpha_3} \cdot \cdots$$ and $$y=+(u \cdot 2^5)^{1/2}$$ implies that the min exponent of $2$ for $y$ would be $$min \left(\frac{2n+1}{2} + \frac{5}{2} \right)$$ or $$ min \left( n+3\right)$$ which is 3 allowing n=0 – MaximusFastidiousIrreverence Sep 05 '17 at 21:42
@AmateurMathGuy See the earlier edits – wlad Sep 17 '17 at 16:17

score 3 · Answer 2 · answered Jun 15 '15 at 19:22

The answer by user3491648 gives a complete characterization of the full set of values $x$ for which $32x+32$ is a square numbers. This is really just an observation on the set of values

$$\{64,1600,53824,1827904,62094400,2109381184\}$$

reported by the OP.

These are not actually values that $x$ takes on. (As user3491648 finds, it requires on odd value of $x$ to make $32x+32$ a square.) Rather, these seem to be the square values that correspond to values of $x$ that are themselves squares, namely of the following values:

$$\{1,7,41,239,1393,8119\}$$

For example, $32\cdot239^2+32=1827904$. It's unclear why these values are being singled out, but they do seem to obey a two-term recursion, namely

$$a_{n+1}=6a_n-a_{n-1}$$

For example, $1393=6\cdot239-41$.

Great observation! I was actually trying to find values for x where (xx)-1 mod 8 == 0 and where 32x*x+32 is a square — Héctor van den Boorn, Jun 15 '15 at 19:28

score -2 · Answer 3 · answered Jun 27 '15 at 04:16

-2

Solution for x such that 32x+32 is a square as follows: For integral solution, x= (n2 -2)/2 such that x is in Natural number.(n is in natural number)

answered Jun 27 '15 at 04:16

Prithwiraj Halder

7

When is $32x+32$ a square number?

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