Number Theory Worksheet 2 – Diophantine Equations
1. Using algebra, find all pairs of positive integers such that both their sum and product are 19.
4
2
1
2. [Source: UKMT Mentoring] Find all integer solutions to 𝑎𝑏 + 𝑎 + 𝑏 = 1.
3. [Source: SMC] For how many integer values of n does the equation x2 + nx − 16 = 0 have
integer solutions?
𝑛−1
4. [Source: SMC] Find all the values of 𝑛 for which both 𝑛 and 4𝑛+1 are integers.
5. For what real values of 𝑥 is 𝑥 2 − 2𝑥 − 2 a perfect square? (i.e. a square number)
6. Each of Paul and Jenny has a whole number of pounds.
He says to her “If you give me £3, I will have 𝑛 times as much as you”.
She says to him: “If you give me £𝑛, I will have 3 times as much as you.
Give that all these statements are true and that 𝑛 is a positive integer, what are the possible
values for 𝑛?
7. [Source: Frosty Special] Find the first five triangular numbers that are perfect squares.
8. Prove that:
a. √4𝑛2 + 1 can never be an integer if 𝑛 is an integer.
b. And hence √𝑥 − √𝑥 is never an integer when 𝑥 is an integer.
(Tip: For equations involving surds, isolate the surd on one side of the equation then
square both sides.)
www.drfrostmaths.com/rzc
Number Theory Worksheet 2 – Diophantine Equations - ANSWERS
1.
Using algebra, find all pairs of positive integers such that their sum and product add up to 19.
Let our two numbers be 𝑎 and 𝑏, and without loss of generality (w.l.o.g.) let 𝑎 ≥ 𝑏.
Using the information, 𝑎𝑏 + 𝑎 + 𝑏 = 19.
Factorising, (𝑎 + 1)(𝑏 + 1) − 1 = 19, so (𝑎 + 1)(𝑏 + 1) = 20.
The possible factor pairs of 20 are 20 × 1, 10 × 2, 5 × 4 (we needn’t consider negative factors in this
particular case, because they’ll lead to negative 𝑎 and 𝑏). This leads to solutions (9,1) and (4,3).
2.
Find all integer solutions to
3.
[Source: SMC] For how many integer values of n does the equation 𝒙𝟐 + 𝒏𝒙 − 𝟏𝟔 = 𝟎 have integer
solutions?
(Official UKMT solution) For the equation to have integer solutions, it must be possible to write 𝑥 2 +
𝑛𝑥 − 16 in the form (x − α) (x − β), where α and β are integers. Therefore 𝑥 2 + 𝑛𝑥 − 16 = 𝑥 2 −
(𝛼 + 𝛽)𝑥 + 𝛼𝛽 and we require that 𝛼𝛽 = −16. The possible integer values of α, β are 1,−16; −1, 16;
2, −8; −2, 8; 4, –4 (we do not count −16, 1 as being distinct from 1, −16, for instance).
As n = −(α + β), the possible values of n are 15, −15, 6, −6 and 0.
4.
Find all the values of 𝒏 for which both 𝒏 and 𝟒𝒏+𝟏 are integers.
𝟒
𝟐
𝟏
+ + = 𝟏.
𝒂
𝒃
4 + 2𝑏 + 𝑎 = 𝑎𝑏
𝑎𝑏 − 2𝑏 − 𝑎 = 4
(𝑎 − 2)(𝑏 − 1) − 2 = 4
(𝑎 − 2)(𝑏 − 1) = 6
Thus we try each of the factor pairs of 6, i.e. -1 and -6, -2 and -3, etc. Thus yields pairs of solutions for
𝑎 and 𝑏 of (1, −5), (0, −2), (−1, −1), (−4,0), (3,7), (4,4), (5,3), (8, 2). But we discard any where 𝑎 or
𝑏 is 0 since we can’t divide by 0 in the original equation.
𝒂𝒃
𝒏−𝟏
𝑛−1
1
𝑛−1
2
𝑛+1
4𝑛+1 is an integer whenever the power is a multiple of , including 0. So let
2𝑛−2
2𝑛+2
4
4
1
= 𝑘. Then
2
rearranging, 𝑘 =
=
−
=2−
.
𝑛+1
𝑛+1
𝑛+1
𝑛+1
The only numbers which divide 4 are -4, -2, -1, 1, 2, 4. This gives values for 𝑛 of −5, −3, −2, 0, 1, 3.
However if 𝑛 = 0 we’d have a negative value of 𝑘. Thus 𝑛 = −5, −3, −2, 1, 3.
5.
For what integer values of 𝒙 is 𝒙𝟐 − 𝟐𝒙 − 𝟐 a perfect square? (i.e. a square number)
𝑥 2 − 2𝑥 − 2 = 𝑘 2
(𝑥 − 1)2 − 3 = 𝑘 2
(𝑥 − 1)2 − 𝑘 2 = 3
(𝑥 + 𝑘 − 1)(𝑥 − 𝑘 − 1) = 3
Note 𝑘 > 0, although 𝑥 may be negative. The only factor pairs of 3 we need to consider are 3 × 1 and
−1 × −3 (since we know 𝑥 + 𝑘 − 1 > 𝑥 − 𝑘 − 1. These gives solutions for 𝑥 of 3 and -1. So there was
indeed a negative solution!
6.
Each of Paul and Jenny has a whole number of pounds.
He says to her “If you give me £3, I will have 𝒏 times as much as you”.
She says to him: “If you give me £𝒏, I will have 3 times as much as you.
Give that all these statements are true and that 𝒏 is a positive integer, what are the possible values
for 𝒏?
Using the information provided, our equations are:
𝑝 + 3 = 𝑛(𝑗 − 3) (1)
𝑗 + 𝑛 = 3(𝑝 − 𝑛) (2)
As per the advice in the lecture slides, if we have three variables, we could use substitution to get a
single equation in terms of two variables. Substituting either 𝑛 or 𝑗 doesn’t yield nice equations we
www.drfrostmaths.com/rzc
can factorise, but it works if we eliminate 𝑝:
𝑝 = 𝑛𝑗 − 3𝑛 − 3 (𝑓𝑟𝑜𝑚 (1))
4𝑛 + 𝑗
𝑝=
(𝑓𝑟𝑜𝑚 (2))
3
4𝑛 + 𝑗
= 𝑛𝑗 − 3𝑛 − 3
3
3𝑛𝑗 − 13𝑛 − 𝑗 = 9
9𝑛𝑗 − 39𝑛 − 3𝑗 = 27
(3𝑛 − 1)(3𝑗 − 13) = 40
Now we consider the factor pairs of 40. Only four lead to integer values for 𝑛 and 𝑗. For example,
using 3𝑛 − 1 = 2 and 3𝑗 − 13 = 20, we get 𝑛 = 1, 𝑗 = 11.
Using our equation above to get 𝑝, we get four possible solutions:
(𝑛, 𝑗, 𝑝) = (1,11,5), (2,7,5), (3, 6,6), (7, 5, 11)
7.
Find the first five triangular numbers that are perfect squares.
1
The nth triangular number is the sum of the first n integers, with the formula 𝑛(𝑛 + 1). Therefore
2
1
2
𝑛(𝑛 + 1) = 𝑘 2 for some 𝑘, and thus 𝑛(𝑛 + 1) = 2𝑘 2 . As discussed in the lecture slides, 𝑛 and 𝑛 + 1
are coprime, and thus either 𝑛 is a square and 𝑛 + 1 is twice a square, or 𝑛 + 1 is a square and 𝑛 is
twice a square. It’s then simply a case of listing out the square numbers, and seeing which we can
either add or subtract one and then half to get a square number (we need only try odd square
numbers, since for even ones, adding or subtracting one gives an odd number, which can’t be halved).
We find this happens for 1 (since half of 2 is a square), 9 (8 is twice a square), 49 (50 is twice a square)
1
and 289 (288 is twice a square). Then using 𝑛(𝑛 + 1) with 𝑛 = 1, 8, 49, 288, we get square
2
triangular numbers of 1, 36, 1225 and 41616.
Side note: I presumed I wasn’t the first person who wondered whether there were square triangular
numbers, so a quick Googling revealed that the problem was studied by Euler, who produced a
formula for generating these numbers, among more general problems. His method involved using a
suitable substitution to yield the Diophantine equation 𝑥 2 − 2𝑦 2 = 1, which is Pell’s equation. See
http://en.wikipedia.org/wiki/Square_triangular_number for more details.
8.
Prove that:
a. √𝟒𝒏𝟐 + 𝟏 can never be an integer if 𝒏 is an integer.
Informally, we could argue that 4𝑛2 is a square number (as √4𝑛2 = 2𝑛), and one more than
a square number is not going to be square itself. We could prove this by showing that the
difference between two square numbers is always greater than 1: (𝑛 + 1)2 − 𝑛2 = 2𝑛 + 1.
When 𝑛 ≥ 1, 2𝑛 + 1 > 1. Thus there can be no two adjacent square numbers.
Note that if we consider 0 a square number, then the above expression would give a square.
b. And hence √𝒙 − √𝒙 is never an integer when 𝒙 is an integer.
Suppose, as a proof by contradiction, that √𝑥 − √𝑥 = 𝑛 where 𝑛 is a positive integer. Then
𝑥 − √𝑥 = 𝑛 2 .
Isolate the √𝑥 on one side of the equation so that we can cleanly square again: 𝑥 − 𝑛2 = √𝑥.
So 𝑥 2 − 2𝑥𝑛2 + 𝑛4 = 𝑥.
Putting this in quadratic form: 𝑥 2 + (−2𝑛2 − 1)𝑥 + 𝑛4 = 0
Using the quadratic formula: 𝑥 =
2n2 +1±√4𝑛2 +1
2
Thus 𝑥 can’t possibly be an integer, because √4𝑛2 + 1 is not an integer.
www.drfrostmaths.com/rzc