You Can Rule Out a Non-Square From Its Last Digit Alone

Look at its last digit. A perfect square can only end in **$0$, $1$, $4$, $5$, $6$ or $9$.

So any number ending in $2$, $3$, $7$ or $8$ is not** a perfect square, and you can say so instantly. $1057$ ends in $7$, so it is not a perfect square — no factorising and no square roots needed.

That single observation rules out four of the ten possible last digits, which makes it the fastest check in the chapter. This page covers the first part of the ICSE Class 8 Mathematics chapter on squares: the properties of square numbers, two methods for squaring quickly, and the patterns that squares form.

A perfect square is a number obtained by multiplying an integer by itself. The first few are

$$1,4,9,16,25,36,49,64,81,100,121,144,169,196,225$$

Property 1 — possible unit digits. A perfect square ends only in **$0$, $1$, $4$, $5$, $6$ or $9$. It never ends in $2$, $3$, $7$ or $8$**.

The reason is that the last digit of a square depends only on the last digit of the number being squared, and squaring the digits $0$ to $9$ produces only those six endings:

- ending in $1$ or $9$ gives a square ending in $1$
- ending in $2$ or $8$ gives a square ending in $4$
- ending in $3$ or $7$ gives a square ending in $9$
- ending in $4$ or $6$ gives a square ending in $6$
- ending in $5$ gives a square ending in $5$
- ending in $0$ gives a square ending in $0$

Property 2 — zeros come in pairs. A perfect square ends in an even number of zeros. So $400$ can be a square (and is, being $20^2$), while $4000$ — with three zeros — cannot be.

Property 3 — even stays even, odd stays odd. The square of an even number is even; the square of an odd number is odd. So $7^2=49$ and $8^2=64$.

Property 4 — squares are never negative, since a positive times a positive and a negative times a negative both give a positive result.

Worked example 1. Is $1057$ a perfect square? It ends in **$7$, so no.

Worked example 2.** Is $4000$ a perfect square? It ends in three zeros, an odd number, so no.

Worked example 3. Is $8649$ a perfect square? It ends in $9$, which is allowed, so the test does not rule it out — and in fact $93^2=8649$.

The limit of the test, which matters. The unit-digit rule can only disprove, never prove. A number ending in $6$ has passed the test but may still not be a square — $26$ ends in $6$ and is not one. So the rule is a quick elimination, and a number that survives it still needs checking properly by prime factorisation.

Getting that direction right is the difference between using the property correctly and misusing it.

Several, and each turns into a shortcut.

Pattern 1 — the sum of consecutive odd numbers. The sum of the **first $n$ odd numbers** is exactly $n^2$:

$$1=1^2$$

$$1+3=4=2^2$$

$$1+3+5=9=3^2$$

$$1+3+5+7+9=25=5^2$$

So the sum of the first $10$ odd numbers is $10^2=100$, with no addition required.

This also gives a test for a perfect square: keep subtracting successive odd numbers $1,3,5,7,\dots$ from a number, and if you reach exactly zero, it is a perfect square, and the count of subtractions is its square root.

Pattern 2 — the difference between consecutive squares. The gap between $n^2$ and the square before it is always an odd number:

$$n^2-(n-1{)}^2=2n-1$$

Checking with $n=8$:

$$8^2-7^2=64-49=15,\text{and }2(8)-1=15$$

This is the same pattern as the first one seen from a different angle: each new odd number added is exactly the gap to the next square.

Pattern 3 — how many numbers sit between two squares. Between $n^2$ and $(n+1{)}^2$ there are exactly **$2n$ numbers that are not** perfect squares.

Checking with $n=4$: between $4^2=16$ and $5^2=25$ lie $17,18,19,20,21,22,23,24$ — that is **$8$** numbers, and $2(4)=8$. Correct.

So the squares get further apart as the numbers grow, which is why perfect squares become rarer the higher you count.

Pattern 4 — Pythagorean triplets. A Pythagorean triplet is a set of three whole numbers $a,b,c$ with $a^2+b^2=c^2$. For any whole number $m>1$, the three numbers

$$2m,m^2-1,m^2+1$$

always form a triplet.

**Worked example — $m=4$:**

$$2m=8,m^2-1=15,m^2+1=17$$

Checking: $8^2+15^2=64+225=289$, and $17^2=289$. Correct.

**Worked example — $m=5$:**

$$10,24,26$$

Checking: $100+576=676=26^2$. Correct.

Other triplets worth recognising on sight: $(3,4,5)$, $(5,12,13)$, $(7,24,25)$ and $(8,15,17)$. Verifying the second: $25+144=169=13^2$.

A limitation of the formula. It generates many triplets but not all of them. Putting $m=2$ gives $(4,3,5)$, which is the familiar $(3,4,5)$ — but $(20,21,29)$ is a genuine triplet that this formula never produces. So the formula is a generator, not a complete description, and a question asking whether three given numbers form a triplet should be answered by **testing $a^2+b^2=c^2$** directly rather than by hunting for an $m$.