Algebra Practice Gym  Anubhav Sehgal, NMIMS Mumbai

How many pairs of nonnegative integers exist, such that the difference between their product and sum is 72?
ab  a  b = 72
ab  a  b + 1 = 73
a(b  1)  1(b  1) = 73
(a  1)(b  1) = 73 = 1 * 73 = 1 * 73
a, b = 2, 74 ; 74, 2 ; 0, 72 ; 72, 0Non negative integer pairs = 1 Unordered or 2 ordered
Total integral solutions = 2 unordered or 4 orderedSince it just says pairs of nonnegative integers without explicitly declaring them as a,b or x,y so we need to take the unordered solutions since the order in which they occur does not hold relevance for this question. Answer: Only 1 i.e. 2, 74.
Find the number of positive integers N for which N^2 + 2014 is square of an integer.
N^2 + 2014 = k^2
k^2  N^2 = 2014
2014 is 4k + 2 form.
So no solutionHow many right angled triangle can be formed of integral side such that one of the three sides is 84?
Case 1: When 84 is not the hypotenuse
x^2 + 84^2 = y^2
x^2  y^2 = 84^2 = 2^4 * 3^2 * 7^2
Number of ways = [f(N/4) – 1]/2
= [f(2^2 * 3^2 * 7^2) – 1]/2
= (27 – 1)/2 = 13.84 can never be hypotenuse of triangle having integral sides
as x^2 + y^2 = 84 = 2^2 * 3 * 7 has (4k + 3) primes with odd powers.
So, 13 ways only.How many integers between 1 and 1000 both inclusive can be expressed as difference of squares of two nonnegative integers?
Case 1: Both numbers are even : 2p, 2q
4p^2  4q^2 = 4(p^2 q^2) = 4k form
Case 2: Both numbers are odd : 2p + 1 , 2q + 1
4p^2 + 4p + 1  4q^2  4q  1 = 4(p^2 + p  q^2  q) = 4k form
Case 3: First even , second odd : 2p, 2q + 1
4p^2  4q^2  4q  1 = 4k  1 = 4k + 3 form
Case 4: First odd , second even : 2p + 1, 2q
4p^2 + 4p + 1  4q^2 = 4k + 1 formHence, difference of squares of two nonnegative integers can be 4k, 4k+1 or 4k + 3 form but never 4k + 2 form
So for 11000 => (3/4)(1000) = 750 cases it can be expressed as the difference of two squares.In how many ways can 7^17 be written as a product of 3 natural numbers?
a * b * c = 7^17
a = 7^x , b = 7^y , c = 7^z
abc = 7^(x + y + z) = 7^17
x + y + z = 17 => C(17 + 3  1,3  1) = C(19,2) non negative integral solutionsBut this includes solutions where any two of a, b, c are equal and permutated in 3!/2! = 3 ways while we need ways to write it as product of 3 natural numbers hence should be counting them only once.
2x + y = 17 => 9 solutions.
Unordering the solutions: (C(19,2)  3*9)/3! + 9 = 33 ways; +9 to include them once.The number of integer solutions of the equation x^2 + 12 = y^4.
x^2 + 12 = y^4
y^4  x^2 = 12
(y^2 + x)(y^2  x) = 12 = 1 * 12 = 2 * 6 = 3 * 4
No solutions for 1 * 12 and 3 * 4 factor pairs as RHS (=12) is even hence we need just even * even factor pairs.
(y^2 + x)(y^2  x) = 2 * 6
y=+/ 2 and x=+/2
4 solutions.For how many positive integers x between 1 and 1000, both inclusive, is 4x^6 + x^3 + 5 divisible by 7?
x^3 mod 7 = 0, 1, 1 only.
Case 1: (4x^6 + x^3 + 5) mod 7 = 5.
Case 2: (4x^6 + x^3 + 5) mod 7 = 4(1) + 1 + 5 mod 7 = 10 mod 7 = 3.
Case 3: (4x^6 + x^3 + 5) mod 7 = 4(1)  1 + 5 mod 7 = 8 mod 7 = 1.
Hence zero values.

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