Prime factorization of $8316$ gives $$8316=2^2\cdot 3^3\cdot 7\cdot 11$$
By setting the equation,
$$8316x=y^2$$
How to find $x$ using prime factorization of $8316$?
Prime factorization of $8316$ gives $$8316=2^2\cdot 3^3\cdot 7\cdot 11$$
By setting the equation,
$$8316x=y^2$$
How to find $x$ using prime factorization of $8316$?
On
$$8316\cdot \underset { x }{ \underbrace { \left( 3\cdot 7\cdot 11 \right) } } =2^{ 2 }\cdot 3^{ 3 }\cdot 7\cdot 11\cdot \left( 3\cdot 7\cdot 11 \right) ={ \left( 2\cdot { 3 }^{ 2 }\cdot 7\cdot 11 \right) }^{ 2 }\quad \\ $$
On
Note that in a prime factorisation, we can immediately see whether a number is a square or not. Taking the square root halves all the exponents, so a number is square exactly when the exponents in its prime factorisation are all even. Now look at the prime factorisation of $8316$ and see what primes there do not appear with even exponents, and use $x$ to fix that.
To be a perfect square, all the prime factors of $y^2$ must be raised to even powers in $y^2$'s prime factorization.
To be the smallest possible $x$ we don't want to add any addition prime factors we don't need.
So $x = 2^a*3^b*7^c*11^d$ and $y = 2^{2e}3^{2f}7^{2g}11^{2h}$ and $8316x = 2^{a+2}3^{b+3}7^{c+1}11^{d+1}=2^{2e}3^{2f}7^{2g}11^{2h}=y$
For $x$ to be as small as possible we want all those variable to be a small as possible.
Example: $a$ must be the smallest number so that $a+2$ is even. That number is 0
$b$ must be the smallest number so that $b +3$ is even. That number is 1.
$c,d$ must be the smallest numbers so that $c+1$ and $d+1$ are even. Those numbers are 1.
So $(2^2*3^3*7*11)(3*7*11) = 2^2*3^4*7^2*11^1 = (2*3^2*7*11)^2$.
So $x = 3*7*11=231$ and $y = 2*3^2*7*11 = 1386$ and $8316*231 = 1386^2$.
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Notice that in general, all the powers of $x$, to be as small as possible but only large enough to make the powers even, that the powers will be $0$ if the original power is even, or $1$ if the original powers are odd.
So the smallest $x$ to make $2^{246}*3^{95391}*5^2*7^{38954}x = y^2$ will be ....$x = 3$. Because $3$ is the only prime factor to an odd power.
$mx = y^2$; $x$ the smallest such number: then $x =$ the product of the prime factors of $m$ that are raised to an odd power.