Number Of Solutions For Equations Involving Difference/Sum Of Perfect Squares - Hemant Malhotra


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    Number of ways in which a natural number can be expressed as difference of two perfect square

    let number = N = x^2 - y^2
    N = (x - y) (x + y)

    Case 1 :

    let N = even number * even number ( product of 2 even number let 4 = 2 * 2)
    so (x - y) (x + y) = even * even
    so x - y = even, x + y = even
    2x = even + even
    so 2x = even (because sum of even + even = even))
    so x = even/2 (which will be an integer)
    so in this case we will get a integral value of x and y

    Case 2 :

    N = even * odd
    x^2 - y^2 = even * odd
    (x - y) (x + y) = even * odd
    x - y = even
    x + y = odd
    2x = even + odd
    2x = odd (because even + odd = odd)
    so x = odd/2 ( odd number /2 will not be an integer so in this case we will not an integer value of x and y)

    Case 3 :

    N = odd * odd
    (x - y)(x + y) = odd * odd
    x - y = odd
    x + y = odd
    2x = odd + odd
    2x = even (because odd + odd = even as 3 + 3 = 6)
    so x = even/2
    which will give integral value

    so This is just basic thing that when number is expressed as even * even or odd * odd then only we could find integral solutions and number could be expressed as difference of two perfect square number

    Type 1 : Number which is a multiple of 4

    Find number of ways in which 20 could be expressed as difference of two perfect square numbers

    Case 1 :

    x^2 - y^2 = 20
    (x - y)(x + y) = 20 = 1 * 20 = 2 * 10 = 4 * 5
    case1 - (x - y) (x + y) = 2 * 10 (even * even)
    so we will find integral value here
    x - y = 2
    x + y = 10
    so x = 6
    y = 4

    Case 2 :

    (x - y) (x + y) = 4 * 5 ( even * odd not useful for us ))
    so total number of positive values of x and y is 1 (6, 4)
    total solution = 4 * positive integral solutions = 4 * 1 = 4
    so total number of solutions = 4

    Why we multiplied by 4 here ???
    x = 6 and y = 4
    so x^2 = 36 and y^2 = 16
    so 36 - 16 = 20
    but if x = -6 and y = 4 then also x^2 = 36 and y^2 = 16
    if x = 6 and y = -4 (same case)
    if x = -6 and y = -4 (same case)
    so there will be 4 possible integral values.

    Direct Formula ( number which is multiple of 4)
    Find the number of positive integer solutions of equation x^2 – y^2 = 20
    solution: (No of factors after dividing by 4)/2.
    Here number = 20 and N/4 = 20/4 = 5 and number of factors of 5 = (1 + 1) = 2
    so ans = 2/2 = 1 ( so positive integer values=1 ) and TOTAL = 4 * 1 = 4

    Type 2 : Number which is a perfect square

    x^2 - y^2 = 16
    (x - y) (x + y) = 16 = 1 * 16 = 2 * 8 = 4 * 4

    Case 1 :
    (x - y) (x + y) = 1 * 16 ( odd * even case rejected)

    Case 2 :
    (x - y)(x + y) = 2 * 8 (even * even case)
    x - y = 2
    x + y = 8
    so x = 5 and y = 3
    (5,3)

    Case 3 :
    (x - y) (x + y) = 4 * 4
    x - y = 4
    x + y = 4
    so x = 4 and y = 0
    (4,0)

    so total positive solutions here = 1 only because (4,0) is not a positive solution

    Now the other values (5,3) (5,-3)(-5,3)(-5,-3) (4,0)(-4,0)
    so total solutions = 6
    Look here total number of solutions are not 4 * positive numbers but 4 * positive + 2

    Direct approach : {(No of factors on the right hand side of given after dividing by 4) – 1 }/2.

    Example:
    x^2 - y^2 = 16
    Divide 16 by 4 = 4
    Number of factors of 4 ( 2^2 ) = 2 + 1 = 3
    NOW ( 3 - 1) / 2 = 1
    Total positive solutions = 1
    Total Integral solutions = 4 * positive + 2 (in case of perfect square)
    so 4 * 1 + 2 = 6 total integral solutions

    Practice question : x^2 - y^2 = 36 find number of total integral solution [ Answer is 6 ]

    Type 3 : N = 4k+2 form

    let number =18 = 4 * 4 + 2 form
    x^2 - y^2 = 18
    (x - y)(x + y) = 1 * 18 = 2 * 9 = 3 * 6
    so all cases are even * odd so there will be no solution for this

    NOTE - Any number of 4k + 2 form can't be expressed as difference of two perfect square
    So x^2 - y^2 = 26 can't be expressed in this form for any integral values of x and y

    Type 4 : When number is Odd

    x^2 - y^2 = 85
    so (x - y)(x + y) = 1 * 85 = 5 * 17
    so odd * odd

    Case 1 :
    (x - y)(x + y) = 1 * 85
    x - y = 1
    x + y = 85
    so x = 43
    y = 42

    Case 2 :
    (x - y)(x + y) = 5 * 17
    x - y = 5
    x + y = 17
    x=11, y=6
    so two positive solutions
    and 2 * 4 = 8 total integral solutions

    Direct Formula : (No of factors) /2

    x^2 - y^2 = 85
    85 = 5 * 17 so factors = 2 * 2 = 4
    so 4/2 = 2 (number of positive integral values =2)
    and total = 2 * 4 = 8

    Practice Questions - Set 1

    1. x^2 - y^2 = 124 ( find total number of integral solutions)
    2. Find number of ways in which 256 could be expressed as difference of two perfect square numbers .
    3. x^2 - y^2 = 21 ( find number of positive integral solutions)
    4. x^2 - y^2 = 122 ( find number of integral solutions)
    5. x^2 - y^2 = 100 ( find the no of non negative integral solution)

    Number of ways in which a natural number can be expressed as Sum of two perfect squares

    Case 1 - when N is not a perfect square and prime factors in the form 4k + 1

    Example : x^2 + y^2 = 5^2 × 13^3
    Here 5 and 13 are both are both of form 4k + 1
    Number of positive integral solutions = Number of Factors of 5^2 × 13^3 = 12
    Total integral solutions = 4 × 12 = 48

    Case 2 : when N is a perfect square

    Example : x^2+y^2= 5^2 ×13^2
    Here number is perfect square and 5 and 13 both are in 4n + 1 form so
    Positive integral solutions = Number of Factors - 1= 9 - 1 = 8
    Total integral solutions= 4 × 8 + 4 = 36 (we will consider only 4k + 1 form of prime factors)

    Extra addition of 4 is for x = +/- 5 * 13 and y = +/- 5 * 13

    Case 3- when x^2 + y^2 = 3^2 × 7^3
    here 3 and 7 are 4k + 3 form and no 4k + 1 form
    So number of integral solutions = zero

    Case 4 - x^2 + y^2 = 5^2 × 3^4
    here 3 is 4k + 3 form and 5 is 4k+1 form
    now 2 cases arise

    1. when 4k + 3 form has odd power then number of integral solutions =0
    2. when 4k + 3 form has even power then ignore that and find number of factors of 4k + 1 form

    Example x^2 + y^2= 5^3 ×7^2
    Here 7 is 4k + 3 form but power is even so ignore that now find factors of 5^3 which is 4.
    Number of positive integral solutions = 4
    and total = 4 × 4 = 16

    Case 5 : When number is of perfect square form and no 4k + 1 form also then
    Number of integral solutions will be 4 only

    Example= x^2 + y^2 = 81 = 3^4
    Here number of integral solutions is 4

    Practice Questions - Set 2

    1. x^2 + y^2 = 80
    2. x^2 + y^2 = 80
    3. x^2 + y^2 = 121
    4. x^2 + y^2 = 126
    5. x^2 + y^2 = 120


  • Set 1

    1. 4
    2. 12
    3. 2
    4. 0


  • @vishwa2017 for question no. 2, 256 is a perfect square, so i think answer should be 14


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