Diophantine Equations - Solve by Bounding Values
Often, one of the most helpful tools in solving Diophantine equations is referred to as "bounding." Frequently, encounters with Diophantine equations can be remarkably vague, often for a reason. This prompts the solver to make assumptions WLOG (without loss of generality) that make the problem easier to handle without fundamentally changing it. Bounding is the process of restricting values of a variable to a manageable set. The best way of illustrating this is through a few examples.
Quadratic Diophantine Equations
General Diophantine Equations
Determine all pairs \((x,y)\) of integers that satisfy the equation
\[(x+1)^4 - (x-1)^4 = y^3.\]
Upon expansion, we reach
\[(x+1)^4 - (x-1)^4 = 8x^3 + 8x.\]
Now, notice that if we assume \(x \geq 1,\)
\[8x^3 = (2x)^3\leq 8x^3 + 8x \leq (2x+1)^3 = 8x^3 + 12x^2 + 6x + 1.\]
From this, we can say that \(8x^3 + 8x\) is between two consecutive perfect (integer) cubes, which is impossible. So, we can say that \(x\) is nonpositive, or \( x \leq 0\). Now, notice that if \((x,y)\) is a solution, then \((-x,-y)\) is also a solution, which implies that \(-x\) is nonpositive, or \(-x \leq 0\). Hence, the only solution is \((0,0)\). \(_\square\)
Solve in positive integers the equation
\[3(xy+yz+xz) = 4xyz.\]
Dividing both sides by \(3xyz\), we reach
\[\frac{1}{z} + \frac{1}{y} + \frac{1}{x} = \frac{4}{3}.\]
Without loss of generality, suppose \(z \leq y \leq x\). This yields
\[\frac{3}{z} \geq \frac{4}{3}.\]
Which implies \(z \leq 2\). Splitting into two separate cases and applying similar logic, we can come up with some systems of equations. Upon solving, the only solutions are \((1,4,12),(1,6,6),(2,2,3)\) and all of their permutations, accounting for the symmetry of the equation. \(_\square\)
Find all pairs of positive integers \((x,y)\) such that
\[x^3-y^3=xy+61.\]
We are given
\[x^3-y^3=(x-y)(x^2+xy+y^2)=xy+61.\]
Notice that \(x>y\). Therefore, we have to consider
\[x^2+xy+y^2≤xy+61 \implies x^2+y^2≤61.\]
Since \(x>y,\) we have
\[61≥x^2+y^2≥2y^2 \implies y\in \{1,2,3,4,5\} \\ \begin{cases} y=1,\quad x^3-x-62=0 \\ y=2,\quad x^3-2x-69=0 \\ y=3,\quad x^3-3x-88=0 \\ y=4,\quad x^3-4x-125=0 \\ y=5, \quad x^3-5x-186=0. \end{cases} \]
Of these equations we can see the only working value of \(x\) is when \(y=5\) and \(x=6\), so the only positive integer pair of solution is \((x,y)=(6,5)\). \(_\square\)