Let c be a positive integer less than 101 such that c=a2−b2
Now, c=a2−b2⇒c=(a+b)(a−b)
Case I : a−b=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 : a−b=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 a−b=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 a−b=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=a2−b2; a,b∈Z
C−I: Difference of a and b is 1.
∴ 12−02=1
22−12=3
32−22=5
42−32=7
52−42=9
.
.
.
502−492=99
All odd number upto 100 can be expressed as difference of two squares.
C−II: Difference of a and b is 2.
22−02=4
32−12=8
42−22=12
.
.
.
262−242=100
All multiplies of '4' upto 100 can be expressed as difference of two squares.
C−III: Difference of a, b is 3
This case will give odd numbers which are already counted
C−IV: 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 C−I and C−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.