Square in Pāṭīgaṇitam

Here we will be knowing how to find square of a number as per Pāṭīgaṇitam.

Verse 23

कृत्वाऽन्त्यपदस्य कृतिं शेषपदैर्द्विगुणमन्त्यमभिहन्यात् ।

उत्सार्य्योत्सार्य पदाच्छेषं चोत्सारयेत् कृतये ॥ २३ ॥

Translation 23

To obtain the square of a number (proceed successively as follows): Having squared the last digit (antya-pada) (i.e., having written the square of the last digit over the last digit), multiply the remaining digits by twice the last, moving it from place to place (towards the right, and set down the resulting products over the respective digits); then (rub out the last digit and) move the remaining digits (one place to the right).

This rule is based on the formula $(a+b)^{2}=a^{2}+2ab+b^{2}$.

This rule is illustrated with an example below.

Example: Square of 125

Here the last digit it 1 . Its square (1² = 1) is placed over itself.

1
1	2	5

Twice the last digit 2 X 1 = 2 placing it below the rest of the digits (2 or 5) and rubbing out the last digit (1). Now the work on the pāṭī appears as

1
	2	5
		2

Multiply 25 by 2 (below) 25 X 2 = 50 and place the result (50) over the respective figures (25), we get

1	5	0
	2	5

One round of operation is completed. Move the remaining digits (25) one place forward to the right. we get

1	5	0
		2	5

Now the process is repeated . In the remaining digits (25) the last digit is 2. Square of the last digit (2² = 4) is placed above 2

1	5	0 + 4
		2	5

1	5	4
		2	5

Twice the last digit 2 X 2 = 4 placing it below the rest of the digit (5) and rubbing out the last digit (2). Now the work on the pāṭī appears as

1	5	4
			5
			4

Multiply 5 by 4 (below) 5 X 4 = 20 and place the result (20) over the respective figure (5), we get

1	5	4 + 2	0
			5

1	5	6	0
			5

Second round of operation is completed. Move the remaining digit (5) one place forward to the right. we get

1	5	6	0
				5

Now the process is repeated . In the remaining digits (5) the last digit is 5. Square of the last digit (5² = 25) is placed above 5

1	5	6	0 + 2	5
				5

1	5	6	2	5
				5

As there are no remaining digits the process ends. 5 being rubbed out. Now the work on the pāṭī appears as

1

5

6

2

5

Square of 125 = 15625

Verse 24

सदृशद्विराशिघातो रूपादिद्विचयपदसमासो (वा) ।

इष्टोनयुतपदवधो वा तदिष्टवर्गान्वितो वर्गः ॥ २४ ॥

Translation 24

The square (of a given number) is also equal to the product of two equal numbers (each equal to the given number), or the sum of as many terms of the series whose first term is 1 and common difference 2, or the product of the difference and the sum of the given number and an assumed number plus the square of the assumed number.

These rules are illustrated with examples below.

The square (of a given number) is also equal to the product of two equal numbers (each equal to the given number)

$n^{2}=n\times n$

Number	Square		Number	Square		Number	Square
1	1 X 1 = 1		4	4 X 4 = 16		7	7 X 7 = 49
2	2 X 2 = 4		5	5 X 5 = 25		8	8 X 8 = 64
3	3 X 3 = 9		6	6 X 6 = 36		9	9 X 9 = 81

The sum of as many terms of the series whose first term is 1 and common difference 2

$n^{2}=1+3+5+7+9+11+13+15+17........+\ to\ n\ terms$

Number	Square		Number	Square		Number	Square
1	1		4	1 + 3 + 5 + 7 = 16		7	1 + 3 + 5 + 7 + 9 + 11 + 13 = 49
2	1 + 3 = 4		5	1 + 3 + 5 + 7 + 9 = 25		8	1 + 3 + 5 + 7 + 9 + 11 + 13 + 15 = 64
3	1 + 3 + 5 = 9		6	1 + 3 + 5 + 7 + 9 + 11 = 36		9	1 + 3 + 5 + 7 + 9 + 11 + 13 + 15 + 17 = 81

The product of the difference and the sum of the given number and an assumed number plus the square of the assumed number

$n^{2}=(n-a)(n+a)+a^{2}$

Let us assume the number $a=1$

Number	Square		Number	Square		Number	Square
1	(1 - 1)(1 + 1) + 12 = 1		4	(4 - 1)(4 + 1) + 12 = 16		7	(7 - 1)(7 + 1) + 12 = 49
2	(2 - 1)(2 + 1) + 12 = 4		5	(5 - 1)(5 + 1) + 12 = 25		8	(8 - 1)(8 + 1) + 12 = 64
3	(3 - 1)(3 + 1) + 12 = 9		6	(6 - 1)(6 + 1) + 12 = 36		9	(9 - 1)(9 + 1) + 12 = 81

See Also

पाटीगणितम् में 'वर्ग'