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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@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 $a\ge b$.

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.

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

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

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

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

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

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, $2|v_p(a)+v_p(b)=v_p(ab). \square$

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

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.