Diophantine Equations: Problems with Solutions

Solution:
Dividing by [tex]2[/tex], we get the equivalent equation
[tex]8x+13y=159[/tex]
Modulo [tex]8[/tex] we have [tex]13\equiv 5[/tex] and [tex]159\equiv 7[/tex], so [tex]5y\equiv 7\pmod 8[/tex]. Since [tex]5\times 5=25\equiv 1\pmod 8[/tex], multiplying by [tex]5[/tex] gives
[tex]y\equiv 35\equiv 3\pmod 8[/tex]
For [tex]y=3[/tex] we get [tex]8x=159-39=120[/tex], i.e. [tex]x=15[/tex], so [tex](x, y)=(15, 3)[/tex] is a particular solution. If [tex](x, y)[/tex] is any solution, subtracting [tex]8\times 15+13\times 3=159[/tex] from [tex]8x+13y=159[/tex] gives
[tex]8(x-15)=-13(y-3)[/tex]
Since [tex]8[/tex] and [tex]13[/tex] are coprime, [tex]y-3[/tex] is divisible by [tex]8[/tex], i.e. [tex]y=3+8t[/tex] with [tex]t\in\mathbb{Z}[/tex]; then [tex]8(x-15)=-104t[/tex], i.e. [tex]x=15-13t[/tex]. Conversely, every such pair is a solution:
[tex]16(15-13t)+26(3+8t)=240-208t+78+208t=318[/tex]
Hence the solutions are [tex]x=15-13t,\ y=3+8t[/tex], where [tex]t\in\mathbb{Z}[/tex].

Solution:
Let [tex]x[/tex] be the number of [tex]130[/tex]-pound crates and [tex]y[/tex] the number of [tex]160[/tex]-pound crates, where [tex]x[/tex] and [tex]y[/tex] are non-negative integers. Then
[tex]130x+160y=3000\iff 13x+16y=300[/tex]
Modulo [tex]13[/tex] we have [tex]16\equiv 3[/tex] and [tex]300=13\times 23+1\equiv 1[/tex], so [tex]3y\equiv 1\pmod{13}[/tex]. Since [tex]3\times 9=27\equiv 1\pmod{13}[/tex], multiplying by [tex]9[/tex] gives
[tex]y\equiv 9\pmod{13}[/tex]
On the other hand, [tex]16y\le 300[/tex], so [tex]y\le 18[/tex], and the only such value is [tex]y=9[/tex]. Then
[tex]13x=300-16\times 9=156\Rightarrow x=12[/tex]
So the truck can be loaded fully in exactly one way: [tex]12[/tex] crates of [tex]130[/tex] pounds and [tex]9[/tex] crates of [tex]160[/tex] pounds ([tex]1560+1440=3000[/tex] pounds). The total number of crates is [tex]12+9=21[/tex].

Solution:
Let the team have [tex]w[/tex] wins, [tex]d[/tex] draws and [tex]l[/tex] losses ([tex]w[/tex], [tex]d[/tex], [tex]l[/tex] are non-negative integers). Then
[tex]w+d+l=38,\qquad 3w+d=80[/tex]
From the second equation [tex]d=80-3w\ge 0[/tex], so [tex]w\le 26[/tex]. Substituting into the first equation, we get
[tex]l=38-w-(80-3w)=2w-42[/tex]
The number of losses grows with [tex]w[/tex], so it is largest for [tex]w=26[/tex]: then [tex]d=2[/tex] and [tex]l=10[/tex]. Check: [tex]26+2+10=38[/tex] matches and [tex]3\times 26+2=80[/tex] points. Hence the largest possible number of losses is [tex]10[/tex].

Solution:
Suppose that integers [tex]x[/tex] and [tex]y[/tex] satisfy the equation. Then [tex]5x^2=11y+7[/tex], so
[tex]5x^2\equiv 7\pmod{11}[/tex]
Since [tex]5\times 9=45\equiv 1\pmod{11}[/tex], multiplying by [tex]9[/tex] gives
[tex]x^2\equiv 63\equiv 8\pmod{11}[/tex]
On the other hand, every integer [tex]x[/tex] is congruent modulo [tex]11[/tex] to one of the numbers [tex]0, \pm 1, \pm 2, \pm 3, \pm 4, \pm 5[/tex], and
[tex]0^2=0,\quad (\pm 1)^2=1,\quad (\pm 2)^2=4,\quad (\pm 3)^2=9,\quad (\pm 4)^2=16\equiv 5,\quad (\pm 5)^2=25\equiv 3\pmod{11}[/tex]
So a perfect square can leave only the remainders [tex]0, 1, 3, 4, 5, 9[/tex] when divided by [tex]11[/tex], and never the remainder [tex]8[/tex]. This contradiction proves that the equation has no solutions in integers.