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Question

Find the number of positive integers less than 101 that cannot be written as the difference of two squares of integers.

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Solution

Let c be a positive integer less than 101 such that c=a2b2
Now, c=a2b2c=(a+b)(ab)

Case I : ab=1
c=a+b
Then, for c to be an integer a+b must be an odd integer.
Hence, c can be an odd integer.

Case II : ab=2
c=2(a+b)
Then, for c to be an integer a+b must be an even integer.
Therefore, c will be a multiple of 4
Hence, c can be a multiple of 4.

Now, if ab=3,5,7,.... any odd integer.
Then, for c to be an integer a+b must be an odd integer, which all have already been counted in Case I

Now, if ab=4,5,7,.... any odd integer.
Then, c will be a multiple of 4, which all have already been counted in Case II

No. of numbers which can be expressed as difference of squares
= (odd numbers) + (multiples of 4) = 50 + 25 = 75
Required numbers which cannot be expressed as a difference of squares of two integers are = 100 - 75 = 25.


Alternate Solution :
Let c be positive integer 100 such that c=a2b2; a,bZ
CI: Difference of a and b is 1.
1202=1
2212=3
3222=5
4232=7
5242=9
.
.
.
502492=99
All odd number upto 100 can be expressed as difference of two squares.

CII: Difference of a and b is 2.
2202=4
3212=8
4222=12
.
.
.
262242=100
All multiplies of '4' upto 100 can be expressed as difference of two squares.

CIII: Difference of a, b is 3
This case will give odd numbers which are already counted

CIV: Difference of a, b is 4
This case will give multiples of 4 which are already counted
Similarly, the remaining cases will all give number already counted in CI and CII
No. of numbers which can be expressed as difference of squares
= (odd numbers) + (multiples of 4) = 50 + 25 = 75
Required numbers which cannot be expressed as a difference of squares of two integers are = 100 - 75 = 25.

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