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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 $8316$?

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    $\begingroup$ Hint: what kinds of powers do squares have? $\endgroup$ Commented Nov 5, 2016 at 19:38
  • $\begingroup$ Smallest positive integer: $2^0\cdot3^1\cdot7^1\cdot11^1$. $\endgroup$ Commented Nov 5, 2016 at 20:19

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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.

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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.

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$$8316\cdot \underset { x }{ \underbrace { \left( 3\cdot 7\cdot 11 \right) } } =2^{ 2 }\cdot 3^{ 3 }\cdot 7\cdot 11\cdot \left( 3\cdot 7\cdot 11 \right) ={ \left( 2\cdot { 3 }^{ 2 }\cdot 7\cdot 11 \right) }^{ 2 }\quad \\ $$

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  • $\begingroup$ I would not call that a hint :-) $\endgroup$
    – TonyK
    Commented Nov 5, 2016 at 19:44
  • $\begingroup$ I guess you're right) $\endgroup$
    – haqnatural
    Commented Nov 5, 2016 at 19:45

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