Theory of Equations  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.Number of positive integral solutions for the equation 1/x + 1/y + 1/z = 23/72 such that x < y < z < 72 are?
Taking the LCM on the LHS as 72, we get, x’ + y’ + z’ = 23 where x’, y’ and z’ will be factors of 72 only.
[Say we had 1/a + 1/b = 5/6 and we took LCM on LHS as 6. We would get [(6/a) + (6/b)]/6 = 5/6 and we could write it as a’ + b’ = 5]
Factors of 72 below 23 = {1, 2, 3, 4, 6, 8, 9, 12, 18}
23 = 18 + 3 + 2 = 18 + 4 + 1 = 12 + 8 + 3 = 12 + 9 + 2
This gives us four sets for (x’, y’, z’) and consequently four sets for x, y, z as:
(72/18, 72/3, 72/2) = 4, 24, 36. Accepted.
(72/18, 72/4, 72/1) = 4, 18, 72. Rejected as z must be less than 72.
(72/12, 72/8, 72/3) = 6, 9, 24. Accepted.
(72/12, 72/9, 72/2) = 6, 8, 36. Accepted.Thus there are 3 valid solution sets for the above given equation
Find the number of positive unequal integral solutions of x1 + x2 + x3 + x4 =20.
First of all, it is important to know that:
x1 + x2 + … + xn = k has
C(k + n – 1, n – 1) nonnegative integral solutions AND
C(k – 1, n – 1) positive integral solutions.Example a + b + c = 6
{(1, 2, 3); (1, 1, 4); (2, 2, 2)} = 3 solutions.
But by formula you get, C (5, 2) = 10.
This is because the formula permutes the solutions as well.
(1, 2, 3) is permuted in 3! = 6 ways.
(1, 1, 4) is permuted in 3C2 = 3 ways.
(2, 2, 2) is permuted in 3C3 = 1 way.
Giving us a total of 10 ways.So here we have, x1 + x2 + x3 + x4 = 20 => C (19, 3) positive integral solutions.
But since there are four variables involved, there will be cases where two, three or all four of them are equal. We need the unequal integral positive solutions.
Consider the case when all four variables are equal 4x1 = 20 => x1 = 5 i.e. 1 solution which will be present as 4C4 = 1 permutation only.
1 * 4C4 = 1 such solutionsNow consider the case when three of the variables are equal
3x1 + x4 = 20
x1 = 1, x4 = 17
…
x1 = 6, x4 = 2
6 solutions which will be present as 4C3 = 4 permutation for each case.
But, note this also includes the case when all four were equal to 5. To avoid double removal, do not consider it here now.
(6 – 1) * 4C3 = 20 such solutionsConsider two equal
2x1 + x3 + x4 = 20.
(72 – 6) * C (4, 2) = 66 * C (4, 2).[How we got 72  6 here ?
2x1 + x3 + x4 = 20
for x1 = 1, x3 + x4 = 18 (17C1 = 17 solutions)
for x1 = 2, x3 + x4 = 16 (15C1 = 15 solutions)
for x1 = 3, x3 + x4 = 14 (13C1 = 13 solutions)
and so on..
for x = 9, x3 + x4 = 2 (1 solution)
So total = 1 + 3 + 5 + ... + 15 + 17 = 81 solutions
Now in all these 9 cases we have a x3 = x4 situation. (Eg: for x1 = 3, x3 + x4 = 14 means x3 = x4 = 7 is possible)
Remove all these 9 cases (where x3 = x4) from the total and we get 81  9 = 72 solutions.
Note this omits the case of all equal which we have already removed earlier. So to avoid double omission.
You can either see it 721 minus 5 (for three equal only)
Or simply (72  6).]So, our answer would be C (19, 3) – 1 * C (4, 4) – (6 – 1)*C (4, 3) * 66 * C (4, 2) = 552 positive unequal integral solutions.
Find the number of positive integral solutions for:
(i) x^2 – 6y^2 = 1340
(ii) x^2 – 16y^2 = 2127(i) x^2 – 6y^2 = 1340
Taking modulus 3 on both sides, we get,
x^2 mod 3 – 6y^2 mod 3 = 2
x^2 mod 3 = 2
But we know that squares are of the form 3k or 3k + 1 form only and cannot be 3k + 2 form.
Hence this equation has no solutions.(ii) x^2 – 16y^2 = 2127
x^2 – (4y)^2 = 2127
(x + 4y) * (x – 4y) = 2127 = P * Q
(x + 4y) = P and (x – 4y) = Q
x = (P + Q)/2
y = (P – Q)/8
Therefore, (P – Q) must be divisible by 8 and (P + Q) must be divisible by 2 for any integral solutions for x, y.2127 = 1 * 2127 = 3 * 709 (2127 + 2) is divisible by 2 but (2127 – 1) = 2126 is not divisible by 8. Hence no integral solutions.
(709 + 3) is divisible by 2 but (709 – 3) = 706 is not divisible by 8. Hence no integral solutions

in the question Find the number of positive unequal integral solutions of x1 + x2 + x3 + x4 =20., how do we get number of Cases for x1=x2 as (726) ?

@test1234
2x1 + x3 + x4 = 20
for x1 = 1, x3 + x4 = 18 (17C1 = 17 solutions)
for x1 = 2, x3 + x4 = 16 (15C1 = 15 solutions)
for x1 = 3, x3 + x4 = 14 (13C1 = 13 solutions)
and so on..
for x = 9, x3 + x4 = 2 (1 solution)
So total = 1 + 3 + 5 + ... + 15 + 17 = 81 solutionsNow in all these 9 cases we have a x3 = x4 situation. (Eg: for x1 = 3, x3 + x4 = 14 means x3 = x4 = 7 is possible)
Remove all these 9 cases (where x3 = x4) from the total and we get 81  9 = 72 solutions.Note this omits the case of all equal which we have already removed earlier. So to avoid double omission.
You can either see it 721 minus 5(for three equal only)
Or simply (72  6).
Hope that clears your doubt.

@anubhav_sehgal Thanks for your reply, it does Clear my doubt. But Can you please do it for odd sum, e.g. x1 + x2 + x3 + x4 =23, it'll be really helpful.

@anubhav_sehgal Could you pls explain this ordered and unordered. how to distinguish it? i know its silly, but no clue about it either.. pls explain
ab  a  b = 72ab  a  b + 1 = 73a(b  1)  1(b  1) = 73(a  1)(b  1) = 73 = 1 * 73 = 1 * 73a, b = 2, 74 ; 74, 2 ; 0, 72 ; 72, 0
Non negative integer pairs = 1 Unordered or 2 ordered
Total integral solutions = 2 unordered or 4 ordered

Consider two equal 2x1 + x3 + x4 = 20.(72 – 6) * C (4, 2) = 66 * C (4, 2).
How is this 726 ? can you pls elaborate

@tanveermalhotra rule of thumb: ordered solutions >= unordered solutions
For unordered, (a,b) and (b,a) are same
while for ordered, (a,b) and (b,a) are different

@tanveermalhotra Check previous question posted by me

@tanveermalhotra already explained by anubhav as a response to @test1234's query and now it is added as part of the article itself.