Diophantine Equations

Does $6x + 9y = 20$ have integer solutions? The left side is always a multiple of $\gcd(6, 9) = 3$, and $20$ is not: none. What about $6x + 9y = 21$? Now $3 \mid 21$, and $x = 2$, $y = 1$ works ($12 + 9 = 21$). With $d = 3$, the general solution is $x = 2 + 3t$, $y = 1 - 2t$.

Find all positive solutions of $7x + 11y = 200$. First solve $7x + 11y = 1$: Euclid gives $7\cdot(-3) + 11\cdot2 = 1$. Multiply by $200$: $x_0 = -600$, $y_0 = 400$. General: $x = -600 + 11t$, $y = 400 - 7t$. Positivity: $11t > 600$ needs $t \ge 55$, and $7t < 400$ needs $t \le 57$. So $t = 55, 56, 57$ give $(5, 15), (16, 8), (27, 1)$.

A farmer spends exactly $\$1000$ on cows at $\$80$ and sheep at $\$30$, buying at least one of each. In how many ways? $80c + 30s = 1000$, or $8c + 3s = 100$. Mod $3$: $8c \equiv 2c \equiv 100 \equiv 1$, so $c \equiv 2 \pmod 3$. Try $c = 2, 5, 8, 11$ (the next, $14$, costs too much): $s = 28, 20, 12, 4$. Four ways.

Solve $xy - 3x - 2y = 7$ in positive integers. The product $(x - 2)(y - 3)$ expands to $xy - 3x - 2y + 6$, so add $6$ to both sides: $(x - 2)(y - 3) = 13$. Since $x \ge 1$ and $y \ge 1$, the factors are at least $-1$ and $-2$, which rules out the negative pairs $(-1, -13)$ and $(-13, -1)$. The pairs $(1, 13)$ and $(13, 1)$ give $(x, y) = (3, 16)$ and $(15, 4)$.

Find all positive integers with $\dfrac1x + \dfrac1y = \dfrac16$. Clear denominators: $6y + 6x = xy$, so $xy - 6x - 6y = 0$ and $(x - 6)(y - 6) = 36$. Could both factors be negative? Then $x - 6$ and $y - 6$ would each be at least $-5$, making the product at most $25 < 36$: impossible. So both are positive divisors of $36$, and $36$ has $9$ of them: $9$ ordered solutions, from $(7, 42)$ to $(42, 7)$ through $(12, 12)$.

Find all positive integers $a \le b \le c$ with $\dfrac1a + \dfrac1b + \dfrac1c = 1$. The largest term is $1/a$, so $1 \le 3/a$ and $a \le 3$; and $a \ge 2$. If $a = 3$, then $1/b + 1/c = 2/3 \le 2/b$ forces $b = 3$, $c = 3$. If $a = 2$, then $1/b + 1/c = 1/2 \le 2/b$ gives $b \le 4$: $b = 3$ gives $c = 6$, $b = 4$ gives $c = 4$. Answer: $(3, 3, 3)$, $(2, 4, 4)$, $(2, 3, 6)$.

Solve $x^2 + y^2 = 2023$. Squares are $0 \pmod 4$ for even numbers and $1$ for odd ones, so $x^2 + y^2 \equiv 0, 1$ or $2 \pmod 4$. But $2023 = 4\cdot505 + 3 \equiv 3$. No solutions — without trying a single pair.

Solve $2^x + 1 = y^2$ in positive integers. Rewrite as $2^x = (y - 1)(y + 1)$: both factors are powers of $2$, and they differ by $2$. The only powers of $2$ two apart are $2$ and $4$, so $y - 1 = 2$, $y = 3$, $x = 3$: indeed $8 + 1 = 9$. Unique factorization did the work — a product equal to a power of $2$ has only powers of $2$ as factors.

Show $x^3 + y^3 + z^3 = 2020$ has no integer solutions. Cubes mod $9$: $n \equiv 0, \pm1, \pm2, \pm3, \pm4$ give $n^3 \equiv 0, \pm1, \pm8, 0, \pm64$, and $\pm8 \equiv \mp1$, $64 \equiv 1$: every cube is $0$ or $\pm1$. Three of them sum to something between $-3$ and $3$, never $\equiv 4$. But $2020 \equiv 2 + 0 + 2 + 0 = 4 \pmod 9$. Impossible.