Square numbers and their properties

A perfect square (or square number) has the form $n^2 = n \times n$ for a whole number $n$. The list starts $1, 4, 9, 16, 25, 36, 49, 64, 81, 100, \dots$ and every entry in it obeys the same set of rules below.

Prime-factorisation test. Write the number as a product of primes. It is a perfect square precisely when every prime's exponent is even — each prime can be split into identical pairs. For $196 = 2^2 \times 7^2$, both exponents are even, so $196$ is a square, and pairing one factor from each pair gives its square root: $2 \times 7 = 14$, so $196 = 14^2$.

Unit-digit filter. A perfect square can only end in $0, 1, 4, 5, 6,$ or $9$; it can never end in $2, 3, 7,$ or $8$. This test only rejects numbers — it never confirms one, since plenty of non-squares also end in an allowed digit.

Zero-count filter. Every zero at the end of a number comes from a hidden factor of $10 = 2 \times 5$. In a perfect square, all prime factors pair up, so the zeros must pair up too — a perfect square always ends in an even number of zeros. A number ending in an odd number of zeros, like $1000$, is automatically ruled out.

Sum-of-odd-numbers pattern. Adding up the first $n$ odd numbers always gives a perfect square:

where: $n$ is how many odd numbers you add, and $(2n-1)$ is the $n$-th odd number itself. For $n=4$: $1+3+5+7 = 16 = 4^2$.

Consecutive-square gap. Two squares that sit next to each other in the list differ by an odd number:

where: $n$ is the smaller of the two whole numbers being squared, and $2n+1$ is the gap between $n^2$ and $(n+1)^2$. For $n=7$: $8^2 - 7^2 = 64-49 = 15 = 2(7)+1$. The picture behind the odd-number pattern is a growing square of dots. A $1\times1$ square is a single dot. To turn it into a $2\times2$ square you wrap an L-shaped border, called a gnomon, of $3$ new dots around it. Wrapping the next gnomon of $5$ dots around the $2\times2$ square gives a $3\times3$ square, and so on — each gnomon is exactly $2$ dots bigger than the last, which is why the running total of gnomons $1, 1+3, 1+3+5, \dots$ always lands on a perfect square.

The gnomon picture can be turned into a short chain of algebra that proves $1+3+5+\dots+(2n-1) = n^2$ for every whole number $n$, not just the small cases you can draw, and it is worth walking through once carefully because the same telescoping idea reappears throughout the chapter whenever a pattern needs a real proof rather than just a few checked examples.

Step 1. Look at what one gnomon contributes. Going from an $(n-1)\times(n-1)$ square of dots to an $n\times n$ square of dots means adding one new column of $n$ dots and one new row of $n$ dots, but the corner dot is shared between them, so the gnomon has $n + n - 1 = 2n-1$ dots — exactly the $n$-th odd number.

Step 2. Write the total after $n$ gnomons as a sum: $S = (2(1)-1) + (2(2)-1) + \dots + (2n-1) = 1+3+5+\dots+(2n-1)$. This is precisely the sum-of-odd-numbers pattern, so proving $S = n^2$ proves the pattern.

Step 3. Use the gnomon meaning directly. Each gnomon takes an $(k-1)\times(k-1)$ square (with $(k-1)^2$ dots) and grows it to a $k\times k$ square (with $k^2$ dots) by adding exactly $2k-1$ dots. So $k^2 - (k-1)^2 = 2k-1$ for every $k$ from $1$ to $n$ (taking $0^2=0$ as the starting point with no dots).

Step 4. Add these differences for $k = 1, 2, \dots, n$. The left side telescopes — each square appears once as an addition and once as a subtraction, cancelling all the middle terms — leaving $n^2 - 0^2 = n^2$. The right side is exactly $1+3+5+\dots+(2n-1)$. So $1+3+5+\dots+(2n-1) = n^2$, confirmed for every $n$, not just the ones you can draw as dots.

Check whether $1444$ is a perfect square using prime factorisation, find its square root, and then verify the result using the consecutive-square gap formula.

Step 1. Unit-digit check: $1444$ ends in $4$, an allowed digit, so it might be a square — this only tells us it is not automatically ruled out.

Step 2. Prime factorise by repeated division: $1444 \div 2 = 722$, $722 \div 2 = 361$, $361 \div 19 = 19$, $19 \div 19 = 1$. So $1444 = 2^2 \times 19^2$ — every exponent is even, confirming $1444$ is a perfect square.

Step 3. Find the square root by pairing: take one factor from each pair, $2 \times 19 = 38$, so $1444 = 38^2$. Check: $38 \times 38 = 1444$.

Step 4. Verify using the gap formula with $n = 38$: the next square should be $39^2 = 38^2 + (2 \times 38 + 1) = 1444 + 77 = 1521$. Direct multiplication confirms $39 \times 39 = 1521$, so the shortcut and the gap formula agree.

$1444 = 38^2$, and the next perfect square after it is $1521 = 39^2$.

• Treating the unit-digit test as proof of a square — it only rejects numbers; $1234$ ends in an allowed digit $4$ but is not a perfect square.
• Miscounting trailing zeros — $1000$ has three zeros (odd), so it cannot be a perfect square, even though it looks "round."
• Forgetting that one odd exponent anywhere in the factorisation, like the $3$ in $2^3 \times 5^2$, breaks the perfect-square condition even if every other prime pairs up.
• Applying the gap formula $(n+1)^2-n^2=2n+1$ across a gap of more than one — it does not give $(n+2)^2 - n^2$ directly; that difference needs its own calculation.

Ask the tutor: Ask the AI tutor to give you a handful of numbers and race the unit-digit, zero-count and factorisation tests against each other to see which spots a non-square fastest.