Show that ab is a perfect square

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Show that ab is a perfect square

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Source: XIII Polish Junior MO 2018 Finals - Problem 1

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#1 h Apr 24, 2018, 1:00 AM • 2 Y

Y by Adventure10, Mango247

Positive odd integers $a, b$ are such that $a^bb^a$ is a perfect square. Show that $ab$ is a perfect square.

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#2 h Apr 24, 2018, 1:32 AM • 8 Y

Y by programjames1, Wizard_32, hansu, mrdriller, NO_SQUARES, howcanicreateacccount, Adventure10, Mango247

WLOG $a \ge b$ ,
We know that $a^{b}b^{a}$ is perfect square.
Then notice that $a^{b}b^{a}=(ab)^{b}b^{a-b}$ . Because $a$ and $b$ are odds, then $a-b$ is even. Therefore, $b^{a-b}$ is perfect square.
Because $a^{b}b^{a}$ is perfect square, then we get $(ab)^{b}$ is perfect square. Because $b$ is odd, then $ab$ is perfect square.

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#3 h Apr 24, 2018, 9:29 AM • 2 Y

Y by Adventure10, Mango247

Redacted...

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Vrangr

#4 h Apr 24, 2018, 10:02 AM • 1 Y

Y by Adventure10

@above, by definition of squares...
$b^{a-b} = (b^{\frac{a-b}{2}})^2$ $\frac{a-b}2$ natural number since, $a - b$ is even and $a\ge b$ .

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#5 h Apr 24, 2018, 10:56 AM • 2 Y

Y by Adventure10, Mango247

khan.academy wrote:

skyerzym27 wrote:

WLOG $a \ge b$ ,
We know that $a^{b}b^{a}$ is perfect square.
Then notice that $a^{b}b^{a}=(ab)^{b}b^{a-b}$ . Because $a$ and $b$ are odds, then $a-b$ is even. Therefore, $b^{a-b}$ is perfect square.
Because $a^{b}b^{a}$ is perfect square, then we get $(ab)^{b}$ is perfect square. Because $b$ is odd, then $ab$ is perfect square.

How $a-b=\text{even}\implies b^{a-b}$ is a perfect square?

The number is said perfect square if and only if that number can be expressed as $x^{2k}$ where $x$ and $k$ are nonnegative integers.

This post has been edited 1 time. Last edited by skyerzym27, Mar 18, 2019, 7:27 AM

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#6 h Apr 24, 2018, 10:56 AM • 2 Y

Y by Adventure10, Mango247

Vrangr wrote:

@above, by definition of squares...
$b^{a-b} = (b^{\frac{a-b}{2}})^2$ $\frac{a-b}2$ natural number since, $a - b$ is odd and $a\ge b$ .

do you mean $\frac{a-b}{2}$ is even

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#7 h Apr 24, 2018, 11:29 AM • 2 Y

Y by Adventure10, Mango247

But then $(b^{\frac{a-b}{2}})^2$ is equal to $b^{2\cdot \frac{a-b}{2}}$ , and we know that (a-b)/2 is an integer so it is indeed a square

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Reason: Wrong latex

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Vrangr

#8 h Apr 24, 2018, 11:40 AM • 2 Y

Y by Adventure10, Mango247

Vrangr wrote:

@above, by definition of squares...
$b^{a-b} = (b^{\frac{a-b}{2}})^2$ $\frac{a-b}2$ natural number since, $a - b$ is even and $a\ge b$ .

skyerzym27 wrote:

do you mean $\frac{a-b}{2}$ is even

Mistyped it earlier

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#9 h Aug 8, 2018, 4:28 AM • 2 Y

Y by Adventure10, Mango247

$a^{b}b^{a}=a^{b-1}b^{a-1}ab$
both b-1 and a-1 is even, so ab is a perfect square

This post has been edited 5 times. Last edited by razor406, Aug 8, 2018, 4:38 AM
Reason: format change

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ATGY

#10 h Jul 23, 2020, 5:48 PM

Y by

Notice that:
$\implies a^b b^a = a^b b^b \cdot b^{(a - b)} = (ab)^b\cdot b^{(a - b)}.$ We know that $a$ and $b$ are odd, hence:
$\implies 2|(a - b).$ This clearly implies that $b^{(a - b)}$ is a perfect square as $b$ is raised to an even power.

For $a^b b^a$ to be a perfect square, we need $(ab)^b$ to be a perfect square as we just showed that $b^{(a - b)}$ is always a perfect square.

It is given in the question that $b$ is odd, which implies that $(ab)^b$ can never be a perfect square, unless $(ab)$ itself is a perfect square.

Thus $(ab)$ must be a perfect square.

Hence proved.

credits to my blog

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#11 h Aug 15, 2020, 6:43 AM • 4 Y

Y by llplp, mrdriller, PNT, trying_to_solve_br

For every prime $p|a^bb^a$ , we have $2| v_p(a^bb^a)=bv_p(a)+av_p(b)$ and since $b,a$ are odd, we must have $v_p(a),v_p(b)$ to be the same parity in order for $a^bb^a$ to a perfect square. Thus, $2|v_p(a)+v_p(b)=v_p(ab). \square$

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lian_the_noob12

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lian_the_noob12

#13 h May 1, 2023, 4:00 PM

Y by

$\color{blue} \boxed{\textbf{SOLUTION}}$

Let, $a=2k+1, b=2k'+1$
So,
$a^{b} b^{a} = (2k+1)^{2k'+1} (2k'+1)^{2k+1}$
$\implies x^2=[(2k+1)^{k'}(2k'+1)^{k}]^{2} (2k+1)(2k'+1)$
$\implies x^2=m^{2} ab$
$\implies ab$ is a perfect square $\blacksquare$

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Amiralizakeri2007

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Amiralizakeri2007

#14 h May 1, 2023, 4:22 PM

Y by

Note that for all $p$ , $bv_p(a)+av_p(b) \equiv 0 (mod2)$ , thus $v_p(a)+v_p(b) \equiv 0 (mod 2)$ $\forall p$ . thus we are done.

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