Square roots by prime factorisation

The square root of a number $x$, written $\sqrt{x}$, is the non-negative number that gives $x$ when multiplied by itself:

where: $x$ is the given perfect square and $n$ is its square root.

Prime factorisation method, in four moves. First, express $x$ as a product of primes using a factor tree or repeated division. Second, group the identical primes into pairs. Third, confirm every prime factor pairs up completely — a perfect square never leaves a prime stranded. Fourth, take one factor from each pair and multiply those together; that product is $\sqrt{x}$. In symbols, if every prime in the factorisation appears to an even power,

where: $p_1, p_2, p_3, \ldots$ are the distinct prime factors of $x$, and $2a, 2b, 2c, \ldots$ are their (even) exponents in $x$.

Since the topic is a reversal of squaring rather than a formula to derive from scratch, the reasoning is the "derivation" here — but it is short and purely structural. Understanding why grouping primes into pairs correctly recovers the square root matters more than memorising the four steps, because it is what lets you catch mistakes and explain the method rather than just execute it, and it is exactly what an exam question testing "explain why" is checking for.

Step 1. Start from what squaring does to a factorisation. Suppose $n = p_1^{a}\times p_2^{b}\times p_3^{c}\times\cdots$ is the prime factorisation of some whole number $n$. Squaring $n$ multiplies the whole expression by itself, and by the laws of exponents that doubles every single exponent: $n^2 = p_1^{2a}\times p_2^{2b}\times p_3^{2c}\times\cdots$.

Step 2. Notice what this guarantees. Every exponent in $n^2$'s factorisation is even, because it is twice a whole number. So every perfect square's prime factorisation has each prime appearing an even number of times — equivalently, each prime factor can be split into pairs with nothing left over.

Step 3. Reverse the process. Given a perfect square $x$, if we factorise it fully and group identical primes into pairs, every pair of $p$'s in $x$'s factorisation corresponds to exactly one $p$ in $n$'s factorisation — because pairing undoes the doubling that squaring caused. Taking one factor from each pair therefore reconstructs $n = \sqrt{x}$ exactly, prime by prime.

Step 4. See why the test direction also works. If, after full factorisation, some prime is left unpaired, its exponent in $x$ is odd — which is impossible for a perfect square, since Step 2 showed every exponent must be even. An unpaired prime is therefore proof that $x$ is not a perfect square, which is why this method also serves as a perfect-square test.

Find $\sqrt{1296}$ by prime factorisation, then use the same number to find the smallest multiplier that turns $1470$ into a perfect square.

Step 1. Factorise $1296$ into primes, dividing by the smallest prime repeatedly.

so $1296 = 2\times2\times2\times2\times3\times3\times3\times3$.

Step 2. Group into pairs: $1296 = (2\times2)\times(2\times2)\times(3\times3)\times(3\times3)$ — four complete pairs, nothing left over, confirming $1296$ is a perfect square.

Step 3. Take one factor from each pair and multiply: $\sqrt{1296} = 2\times2\times3\times3 = 4\times9 = \mathbf{36}$. Check: $36^2 = 1296$. ✓

Step 4. Now apply the same pairing test to a number that is not yet a perfect square. Factorise $1470 = 2\times3\times5\times7\times7$. Here $2$, $3$ and $5$ each appear only once (unpaired), while $7\times7$ is already a complete pair. To make every prime pair up we need one more $2$, one more $3$ and one more $5$, so the smallest multiplier is $2\times3\times5 = 30$.

Step 5. Multiplying gives $1470\times30 = 44100 = (2\times2)\times(3\times3)\times(5\times5)\times(7\times7)$, and taking one factor from each pair, $\sqrt{44100} = 2\times3\times5\times7 = \mathbf{210}$. Check: $210^2 = 44100$. ✓

• Stopping the factor tree too early — leaving a factor like $12$ un-broken instead of continuing to $2\times2\times3$ hides whether the primes actually pair up.
• Taking two factors from a pair (writing $2\times2$ instead of just $2$) when building the final square root, instead of the correct one factor per pair.
• Assuming a number is a perfect square just because it looks factorisable — always check that every prime is paired with none left over before concluding it is one.
• Forgetting to multiply the "one from each pair" values together at the end, and listing the primes separately as if that were the answer.
• Mis-copying a repeated prime factor as two unrelated numbers — for example writing $49 = 7\times7$ but then treating the two $7$s as separate primes when pairing, which throws off the whole count.

Ask the tutor: Stuck factorising a number into primes, or not sure why an unpaired prime rules out a perfect square — ask the AI tutor to walk through your number step by step.