Why Can a Perfect Square End Only in 0, 1, 4, 5, 6, or 9?

A quick way to rule out a number as a perfect square is to inspect its final digit. In ordinary base-10 notation, no integer square ends in 2, 3, 7, or 8. The reason is that the last digit of a square is determined entirely by the last digit of the integer being squared.

Why only the last digit matters

Since $n^2=(-n)^2$, it is enough to consider nonnegative integers. Any nonnegative integer $n$ can be written as $n=10q+r$, where $q$ is an integer and $r$ is one of $0,1,\ldots,9$. The remainder $r$ is the last digit of $n$. Squaring gives

$$(10q+r)^2=100q^2+20qr+r^2.$$

The first two terms are multiples of 10, so they cannot change the final digit. Equivalently, modulo 10,

$$n^2\equiv r^2\pmod{10}.$$

Thus, to find every possible final digit of a square, it is enough to square the ten possible final digits of an integer.

Last digit of $n$	Last digit of $n^2$
0	0
1	1
2	4
3	9
4	6
5	5
6	6
7	9
8	4
9	1

The only resulting digits are $0,1,4,5,6,$ and $9$. Therefore, if an integer ends in $2,3,7,$ or $8$, it cannot be a perfect square.

A direct proof by cases

You can also group the possible last digits into pairs that give the same square ending:

• If $n$ ends in 0, then $n^2$ ends in 0.
• If $n$ ends in 1 or 9, then $n^2$ ends in 1.
• If $n$ ends in 2 or 8, then $n^2$ ends in 4.
• If $n$ ends in 3 or 7, then $n^2$ ends in 9.
• If $n$ ends in 4 or 6, then $n^2$ ends in 6.
• If $n$ ends in 5, then $n^2$ ends in 5.

These cases cover every possible integer last digit and establish the claim. The argument works for negative integers too: a negative integer still has one of the same ten final digits, and its square is nonnegative.

Worked examples

Consider $137^2$. The number 137 ends in 7, so its square must end in 9. Indeed,

$$137^2=18{,}769.$$

For another example, 24 ends in 4, so its square must end in 6. In fact,

$$24^2=576.$$

The test is especially useful for rejecting candidates without calculating a square root. For example, 1,238 cannot be a perfect square because it ends in 8. There is no need to test nearby squares.

The test is necessary, not sufficient

A perfect square must end in one of the six listed digits, but not every number with one of those endings is a square. For instance, 21 ends in 1, an allowed square ending, but it is not a perfect square: $4^2=16<21<25=5^2$.

So the last-digit test has a one-way use: an ending of 2, 3, 7, or 8 proves that a number is not a square; an ending of 0, 1, 4, 5, 6, or 9 merely means the test does not rule it out. Other methods, such as checking whether the integer square root is an integer, are needed to decide whether it actually is a square.

Common mistakes

• Treating the list as a characterization. The list gives possible endings, not a guarantee that every number with one of those endings is square.
• Using the wrong base. This exact list applies to the usual base-10 notation. In another base, the possible final digits are determined by squares modulo that base and may differ.
• Checking more than the last digit when only this test is needed. The final digit of $n^2$ depends only on the final digit of $n$, because every other term in $(10q+r)^2$ is divisible by 10.