Diophantine equations

A Diophantine equation is any equation of the form

$$f(x_1, x_2, x_3, \dots, x_n) = 0$$

where:

• $f$ is a polynomial with integer coefficients,

• the variables $x_1, x_2, \dots, x_n$ are required to take integer values,

• and usually $n \geq 1$ (not necessarily $n \geq 2$).

If $f$ is polynomial with integer coefficients, then it is called an algebraic Diophantine equation.

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Ex:
Determine the number of integers 'n' for which n^2 + 19n + 92 is square.
Let n^2 + 19n + 92 = k^2
n^2 + 2.(19/2).n + 92 = k^2
(19/2)^2 = 361/4
92 - 361/4 = 7/4
=> n^2 + 2.(19/2).n + 361/4 + 7/4 = k^2
=> (n + 19/2)^2 = k^2 - 7/4
=> (2n + 19)^2 = 4k^2 - 7
=> 4k^2 - (2n + 19)^2 = 7
=> (|2k| + |2n + 19|)(|2k| - |2n + 19|) = 7
First factor is bigger so:
|2k| + |2n + 19| = 7
|2k| - |2n + 19| = 1
Add to get:
|2k| = 4 and |2n + 19| = 3
=> 2n + 19 = -+3
=> 2n = -19 -+ 3 => n = -11,-8 = Answer

Q. Find all unordered pairs of natural number, the difference of whose square is 45.
Solution:
a^2 - b^2 = 45
(a + b)(a - b) = 45
Here a,b are natural so no need to take modulus like the last question.
a+b is the bigger factor, so:
a + b = 45, a - b = 1 => a = 23, b = 22
a + b = 15, a - b = 3 => a = 9, b = 6
a + b = 9, a - b = 5 => a = 7, b = 2
Rest of the solutions would just flip a,b and the question says unordered so we just have these 3 answers.

Ex. Find all positive integers 'n' such that n^2 + 96 = perfect square.
Solution:
k^2 - n^2 = 96
(|k| + n)(|k| - n) = 96
Factor pair: 96,1 no solution as k,n should be integers.
Factor pair: 48,2 n = 23.
Factor pair: 32,3 not integer.
Factor pair: 24,4 n = 10.
Factor pair: 16,6 n = 5.
Total 4 answers.

Ex. 

Find all the positive integers $x, y, z$ satisfying

$$x^{\,y^{z}} \cdot y^{\,z^{x}} \cdot z^{\,x^{y}} = 5xyz.$$

Solution:
Divide both sides by xyz to get:
x^(y^z - 1).y^(z^x - 1).z^(x^y - 1) = 5
Since every term is integer => there are 3 ways to distribute integer factors across these 3 terms:
5,1,1
1,5,1
1,1,5
We can solve for one case and rest will be similar.
x^(y^z - 1) = 5 => x = 5, y^z - 1 = 1 => y^z = 2 => y = 2, z = 1
x = 5, y = 2, z = 1 which can be verified by putting these in the original equation.
Similarly other solutions will be:
y = 5, z = 2, x =1
z = 5, x = 2, y = 1.

Ex:

$$\sqrt{x} + y = 7$$ $$x + \sqrt{y} = 11$$

Find the natural numbers $x,y$

Solution:
x and y have to be perfect squares going by the given equations.
Let x = a^2, y = b^2.
a + b^2 = 7
a^2 + b = 11
Subtract:
a^2 - b^2 + b - a = 4
(a-b)(a + b - 1) = 4
Since both factors on LHS are integers, we can assign them factors of 4.
And a + b - 1 >= a - b so the first term will be having the bigger factor.
Case 1:
a - b = 1
a + b - 1 = 4
2a = 6 => a = 3 => b = 2

Case 2:

a - b = 2
a + b - 1 = 2
2a = 5 => a not integer. Ignore.
Answer: a = 3, b = 2

Ex: What is the largest n for which 4^27 + 4^1000 + 4^n is a perfect square?
Solution:

Answer: $n = 1972$

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1. Rewrite everything in powers of 2

$$4^{27}+4^{1000}+4^{n} =\;2^{54}+2^{2000}+2^{2n}\;.$$

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2. Factor out the smallest power of 2

Write

$$2^{54}+2^{2000}+2^{2n} =2^{54}\Bigl(1+2^{1946}+2^{2n-54}\Bigr).$$

Because $2^{54}$ itself is a perfect square, the whole expression is a square iff

$$B \;=\;1+2^{1946}+2^{2n-54}$$

is a perfect square.

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3. Use the only way three powers of 2 plus 1 form a square

A classical (and easily proved) identity tells us

$$\bigl(2^{k}+1\bigr)^2 \;=\;1+2^{k+1}+2^{2k}.$$

Thus the only chance for $B$ to be a square is to match
$1+2^{k+1}+2^{2k}$.

So we must have the set of exponents $\{\,0,\;k+1,\;2k\,\}$ equal to
$\{\,0,\;1946,\;2n-54\,\}$.

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4. Match the exponents — two possible cases

Which exponent is largest?	Conditions that arise	Value of $n$
$2^{1946}$ largest ($2n-54\le 1946$)	$2k=1946\;\Rightarrow\;k=973$ and $k+1 = 2n-54$	$2n-54 = 974 \;\Rightarrow\; n=514$
$2^{2n-54}$ largest ($2n-54\ge 1946$)	$2k = 2n-54 \;\Rightarrow\; k = n-27$ and $k+1 = 1946$	$n-27 = 1945 \;\Rightarrow\; n=1972$

No other match is possible, so the only admissible $n$ are $514$ and $1972$.

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5. Take the largest

$$\boxed{n=1972}.$$

(A quick check: with $n=1972$ the bracket becomes $1+2^{1946}+2^{3890}=(2^{1945}+1)^2$, exactly a perfect square, confirming the result.)