Diophantine Equations: Difficult Problems with Solutions

Solution:
The equation has a solution exactly when the number [tex]n=c-3[/tex] can be written as [tex]n=5a+7b[/tex] with integers [tex]a,b\ge 0[/tex]. For [tex]n=23[/tex], i.e. [tex]c=26[/tex], we need [tex]7b\le 23[/tex], so [tex]b\in\{0, 1, 2, 3\}[/tex] and
[tex]23-7b\in\{23, 16, 9, 2\}[/tex]
None of these numbers is divisible by [tex]5[/tex], so for [tex]c=26[/tex] the equation has no solution.
Now let [tex]n\ge 24[/tex]. The numbers [tex]7b[/tex] for [tex]b=0, 1, 2, 3, 4[/tex] give all possible remainders modulo [tex]5[/tex]:
[tex]0\equiv 0,\quad 7\equiv 2,\quad 14\equiv 4,\quad 21\equiv 1,\quad 28\equiv 3\pmod 5[/tex]
Choose [tex]b\in\{0, 1, 2, 3, 4\}[/tex] with [tex]7b\equiv n\pmod 5[/tex]. If [tex]b\le 3[/tex], then [tex]7b\le 21\lt n[/tex]. If [tex]b=4[/tex], then [tex]n\equiv 3\pmod 5[/tex], and together with [tex]n\ge 24[/tex] this gives [tex]n\ge 28=7b[/tex]. In both cases [tex]n-7b[/tex] is a non-negative multiple of [tex]5[/tex], i.e. [tex]n-7b=5a[/tex] with [tex]a\ge 0[/tex].
Therefore the equation has a solution for every [tex]c\ge 27[/tex], and the largest [tex]c[/tex] for which it has no solution is [tex]c=26[/tex].

Solution:
Write the equation as
[tex](x-y)(x+y)=400[/tex]
The numbers [tex]x-y[/tex] and [tex]x+y[/tex] have the same parity, because their sum [tex]2x[/tex] is even; their product is even, so both are even. Put [tex]x-y=2u[/tex], [tex]x+y=2v[/tex] with integers [tex]u[/tex] and [tex]v[/tex]; then
[tex]uv=100,\qquad x=u+v,\quad y=v-u[/tex]
Conversely, every ordered pair of integers [tex](u, v)[/tex] with [tex]uv=100[/tex] gives the solution [tex](x, y)=(u+v, v-u)[/tex], and different pairs give different solutions, since [tex]u=\frac{x-y}{2}[/tex] and [tex]v=\frac{x+y}{2}[/tex] are determined by [tex](x, y)[/tex]. So we have to count the ordered integer pairs [tex](u, v)[/tex] with [tex]uv=100[/tex].
The number [tex]100=2^2\times 5^2[/tex] has [tex](2+1)(2+1)=9[/tex] positive divisors, so there are [tex]9[/tex] such pairs with [tex]u,v\gt 0[/tex] and [tex]9[/tex] pairs with [tex]u,v\lt 0[/tex]. Hence the equation has [tex]18[/tex] integer solutions (with all combinations of signs):
[tex](\pm 20, 0),\ (\pm 25, \pm 15),\ (\pm 29, \pm 21),\ (\pm 52, \pm 48),\ (\pm 101, \pm 99)[/tex]

Solution:
Collect the terms containing [tex]x[/tex]:
[tex]x(y-3)=2y^2+1[/tex]
If [tex]y=3[/tex], we get [tex]0=19[/tex], which is false. Hence [tex]y\ne 3[/tex] and, since [tex]2y^2+1=(y-3)(2y+6)+19[/tex],
[tex]x=\frac{2y^2+1}{y-3}=2y+6+\frac{19}{y-3}[/tex]
Thus [tex]x[/tex] is an integer exactly when [tex]y-3[/tex] divides [tex]19[/tex], i.e. [tex]y-3\in\{1, -1, 19, -19\}[/tex]:
[tex]y-3=1\Rightarrow y=4,\ x=8+6+19=33[/tex]
[tex]y-3=-1\Rightarrow y=2,\ x=4+6-19=-9[/tex]
[tex]y-3=19\Rightarrow y=22,\ x=44+6+1=51[/tex]
[tex]y-3=-19\Rightarrow y=-16,\ x=-32+6-1=-27[/tex]
All four pairs satisfy the equation (for example, [tex]33\times 4=132=2\times 16+99+1[/tex] and [tex](-27)(-16)=432=2\times 256-81+1[/tex]), so the solutions are [tex](33, 4)[/tex], [tex](-9, 2)[/tex], [tex](51, 22)[/tex] and [tex](-27, -16)[/tex].

Solution:
Each factor is greater than [tex]1[/tex], and the factor [tex]1+\frac{1}{n}[/tex] decreases as [tex]n[/tex] grows. If [tex]a=1[/tex], the first factor already equals [tex]2[/tex] and the product is greater than [tex]2[/tex]. If [tex]a\ge 4[/tex], then [tex]b,c\ge 4[/tex] and
[tex]\left(1+\frac{1}{a}\right)\left(1+\frac{1}{b}\right)\left(1+\frac{1}{c}\right)\le\left(\frac{5}{4}\right)^3=\frac{125}{64}\lt 2[/tex]
Hence [tex]a=2[/tex] or [tex]a=3[/tex].
Case [tex]a=2[/tex]. Then [tex]\left(1+\frac{1}{b}\right)\left(1+\frac{1}{c}\right)=\frac{4}{3}[/tex], i.e. [tex]3(b+1)(c+1)=4bc[/tex], which is equivalent to
[tex]bc-3b-3c-3=0\iff (b-3)(c-3)=12[/tex]
Since the second factor is greater than [tex]1[/tex], we have [tex]1+\frac{1}{b}\lt\frac{4}{3}[/tex], i.e. [tex]b\gt 3[/tex]. So [tex]1\le b-3\le c-3[/tex] and [tex](b-3, c-3)\in\{(1, 12),\ (2, 6),\ (3, 4)\}[/tex], which gives the triples [tex](2, 4, 15)[/tex], [tex](2, 5, 9)[/tex], [tex](2, 6, 7)[/tex].
Case [tex]a=3[/tex]. Then [tex]\left(1+\frac{1}{b}\right)\left(1+\frac{1}{c}\right)=\frac{3}{2}[/tex], i.e. [tex]2(b+1)(c+1)=3bc[/tex], which is equivalent to
[tex]bc-2b-2c-2=0\iff (b-2)(c-2)=6[/tex]
Here [tex]b\ge a=3[/tex], so [tex]1\le b-2\le c-2[/tex] and [tex](b-2, c-2)\in\{(1, 6),\ (2, 3)\}[/tex], which gives the triples [tex](3, 3, 8)[/tex] and [tex](3, 4, 5)[/tex].
All five triples satisfy [tex]a\le b\le c[/tex], so there are [tex]5[/tex] such triples.

Solution:
Adding twice the second equation to the first, we get
[tex](x+y+z)^2=138+2\times 131=400\Rightarrow x+y+z=\pm 20[/tex]
Subtracting the second equation from the first and doubling, we get
[tex](x-y)^2+(y-z)^2+(z-x)^2=2\times 138-2\times 131=14[/tex]
Each of these squares is at most [tex]14[/tex], so it is one of [tex]0, 1, 4, 9[/tex], and the only way to write [tex]14[/tex] as a sum of three such squares is [tex]9+4+1[/tex]. So the three numbers are pairwise different, and when they are arranged in increasing order, the two gaps between neighbours are [tex]1[/tex] and [tex]2[/tex] in some order (the largest difference is [tex]3[/tex]). Thus the numbers are [tex]t,\ t+1,\ t+3[/tex] or [tex]t,\ t+2,\ t+3[/tex] for some integer [tex]t[/tex].
In the first case the sum is [tex]3t+4=\pm 20[/tex], which gives [tex]t=-8[/tex] (the value [tex]t=\frac{16}{3}[/tex] is not an integer): the numbers [tex]-8, -7, -5[/tex]. In the second case [tex]3t+5=\pm 20[/tex] gives [tex]t=5[/tex] (the value [tex]t=-\frac{25}{3}[/tex] is not an integer): the numbers [tex]5, 7, 8[/tex].
Check: [tex]25+49+64=138[/tex] and [tex]35+56+40=131[/tex]; for [tex]-8, -7, -5[/tex] the values are the same. Both equations are symmetric, so every order of each of the two sets of numbers is a solution, and the number of ordered triples is
[tex]2\times 3!=12[/tex]

Solution:
If [tex]n=1[/tex], then [tex]3^m=3[/tex] and [tex]m=1[/tex]. Let now [tex]n\ge 2[/tex]. Then [tex]3^m=1+2^n[/tex] gives remainder [tex]1[/tex] when divided by [tex]4[/tex], while [tex]3\equiv -1\pmod 4[/tex]:
[tex]3^m\equiv (-1)^m\equiv 1\pmod 4[/tex]
so [tex]m[/tex] is even: [tex]m=2k[/tex], [tex]k\ge 1[/tex]. Then
[tex]2^n=3^{2k}-1=\left(3^k-1\right)\left(3^k+1\right)[/tex]
Both factors are positive divisors of a power of [tex]2[/tex], hence powers of [tex]2[/tex] themselves, and they differ by [tex]2[/tex]. The smaller one is [tex]3^k-1\ge 2[/tex]. If it were at least [tex]4[/tex], both factors would be divisible by [tex]4[/tex], and so would their difference [tex]2[/tex], which is false. Hence
[tex]3^k-1=2\Rightarrow k=1,\quad m=2,\quad 2^n=2\times 4=8,\quad n=3[/tex]
Check: [tex]3^1-2^1=1[/tex] and [tex]3^2-2^3=1[/tex]. The solutions are [tex](1, 1)[/tex] and [tex](2, 3)[/tex].

Solution:
We compare the right-hand side with the neighbouring cubes [tex]y^3[/tex] and [tex](y+1)^3[/tex]:
[tex]x^3-y^3=2y^2+1\gt 0[/tex]
[tex](y+1)^3-x^3=\left(y^3+3y^2+3y+1\right)-\left(y^3+2y^2+1\right)=y^2+3y=y(y+3)[/tex]
If [tex]y\ge 1[/tex] or [tex]y\le -4[/tex], then [tex]y(y+3)\gt 0[/tex], so [tex]y^3\lt x^3\lt (y+1)^3[/tex], i.e. [tex]y\lt x\lt y+1[/tex], which is impossible for an integer [tex]x[/tex]. It remains to check [tex]y\in\{-3, -2, -1, 0\}[/tex]:
[tex]y=-3:\quad x^3=-27+18+1=-8\Rightarrow x=-2[/tex]
[tex]y=-2:\quad x^3=-8+8+1=1\Rightarrow x=1[/tex]
[tex]y=-1[/tex]: [tex]x^3=-1+2+1=2[/tex], which has no integer solution;
[tex]y=0:\quad x^3=1\Rightarrow x=1[/tex]
The solutions are [tex](1, 0)[/tex], [tex](1, -2)[/tex] and [tex](-2, -3)[/tex].