Diophantine Equations: Very Difficult Problems with Solutions

Solution:
For [tex]p=2[/tex] we would need [tex]2x^2=3[/tex], which is impossible, so [tex]p[/tex] is an odd prime. Moreover,
[tex]2y^2=p^2+1\gt p+1=2x^2\Rightarrow y\gt x,\qquad 2y^2=p^2+1\lt 2p^2\Rightarrow y\lt p[/tex]
Subtracting the first equation from the second:
[tex]2\left(y^2-x^2\right)=p^2-p\iff 2(y-x)(y+x)=p(p-1)[/tex]
The odd prime [tex]p[/tex] divides [tex]2(y-x)(y+x)[/tex], so it divides [tex]y-x[/tex] or [tex]y+x[/tex]. But [tex]0\lt y-x\lt p[/tex], so [tex]p\mid x+y[/tex]; and since [tex]0\lt x+y\lt 2p[/tex], we get [tex]x+y=p[/tex]. Then [tex]2(y-x)p=p(p-1)[/tex] gives [tex]y-x=\frac{p-1}{2}[/tex] and
[tex]2x=(x+y)-(y-x)=p-\frac{p-1}{2}=\frac{p+1}{2}\Rightarrow p+1=4x[/tex]
Comparing with [tex]p+1=2x^2[/tex] and using [tex]x\gt 0[/tex], we get
[tex]2x^2=4x\Rightarrow x=2,\quad p=7,\quad y=p-x=5[/tex]
Check: [tex]7+1=2\times 2^2[/tex] and [tex]7^2+1=2\times 5^2[/tex]. The answer is [tex]p=7[/tex].

Solution:
For [tex]n=1[/tex] the sum equals [tex]100\ne 2009[/tex], so [tex]n\ge 2[/tex]. Since [tex]n\equiv 1\pmod{n-1}[/tex], every power [tex]n^{x_i}[/tex] (including [tex]n^0=1[/tex]) satisfies [tex]n^{x_i}\equiv 1\pmod{n-1}[/tex], so the left-hand side is congruent to [tex]100[/tex] modulo [tex]n-1[/tex]. Hence
[tex]n-1\mid 2009-100=1909=23\times 83\Rightarrow n-1\in\{1, 23, 83, 1909\}\Rightarrow n\in\{2, 24, 84, 1910\}[/tex]
All four values work. For [tex]n=24[/tex], [tex]84[/tex] and [tex]1910[/tex] it suffices to take terms [tex]n^1=n[/tex] and [tex]n^0=1[/tex]:
[tex]2009=83\times 24+17\times 1,\qquad 2009=23\times 84+77\times 1,\qquad 2009=1\times 1910+99\times 1[/tex]
with [tex]83+17=23+77=1+99=100[/tex] terms in each case. For [tex]n=2[/tex] the binary representation
[tex]2009=2^{10}+2^9+2^8+2^7+2^6+2^4+2^3+2^0[/tex]
uses [tex]8[/tex] terms. Replacing a term [tex]2^k[/tex], where [tex]k\ge 1[/tex], by two terms [tex]2^{k-1}[/tex] keeps the sum and increases the number of terms by [tex]1[/tex]. This can be repeated as long as some term is greater than [tex]1[/tex], i.e. up to [tex]2009[/tex] terms, so in particular exactly [tex]100[/tex] terms can be reached.
Therefore [tex]n\in\{2, 24, 84, 1910\}[/tex].

Solution:
Let the sides be [tex]a,b,c[/tex] and the semiperimeter [tex]s[/tex], and put [tex]x=s-a[/tex], [tex]y=s-b[/tex], [tex]z=s-c[/tex]. By the triangle inequality these numbers are positive, and
[tex]x+y+z=3s-2s=s,\qquad a=y+z,\quad b=x+z,\quad c=x+y[/tex]
By Heron's formula the area is [tex]S=\sqrt{sxyz}[/tex], and the perimeter is [tex]2s[/tex]. The condition [tex]S=2s[/tex] gives
[tex]sxyz=4s^2\iff xyz=4s=4(x+y+z)[/tex]
The numbers [tex]2x=b+c-a[/tex], [tex]2y=a+c-b[/tex], [tex]2z=a+b-c[/tex] have the same parity as the perimeter [tex]a+b+c[/tex]. If the perimeter were odd, then [tex]X=2x[/tex], [tex]Y=2y[/tex], [tex]Z=2z[/tex] would be odd integers with [tex]\frac{XYZ}{8}=4\times \frac{X+Y+Z}{2}[/tex], i.e. [tex]XYZ=16(X+Y+Z)[/tex], an odd number equal to an even one. Hence the perimeter is even, and [tex]x,y,z[/tex] are positive integers with [tex]xyz=4(x+y+z)[/tex].
Let [tex]x\ge y\ge z[/tex]. Then [tex]xyz=4(x+y+z)\le 12x[/tex], so [tex]yz\le 12[/tex]. Moreover, [tex]x(yz-4)=4(y+z)\gt 0[/tex], so [tex]yz\gt 4[/tex] and
[tex]x=\frac{4(y+z)}{yz-4}[/tex]
If [tex]z=1[/tex], then [tex]5\le y\le 12[/tex], and [tex]x=\frac{4(y+1)}{y-4}[/tex] is an integer with [tex]x\ge y[/tex] only for [tex]y=5[/tex] ([tex]x=24[/tex]), [tex]y=6[/tex] ([tex]x=14[/tex]) and [tex]y=8[/tex] ([tex]x=9[/tex]).
If [tex]z=2[/tex], then [tex]3\le y\le 6[/tex] and [tex]x=\frac{2(y+2)}{y-2}[/tex]: [tex]y=3[/tex] gives [tex]x=10[/tex], [tex]y=4[/tex] gives [tex]x=6[/tex], [tex]y=5[/tex] gives a non-integer, and [tex]y=6[/tex] gives [tex]x=4\lt y[/tex].
If [tex]z=3[/tex], then [tex]y\in\{3, 4\}[/tex] and [tex]x=\frac{24}{5}[/tex] or [tex]x=\frac{28}{8}[/tex], which are not integers. If [tex]z\ge 4[/tex], then [tex]yz\ge 16\gt 12[/tex].
The five triples [tex](x, y, z)[/tex] give the triangles with sides [tex](a, b, c)=(y+z, x+z, x+y)[/tex]:
[tex](24, 5, 1)\to(6, 25, 29),\qquad (14, 6, 1)\to(7, 15, 20),\qquad (9, 8, 1)\to(9, 10, 17)[/tex]
[tex](10, 3, 2)\to(5, 12, 13),\qquad (6, 4, 2)\to(6, 8, 10)[/tex]
All the steps are reversible, so all five triangles are perfect; for example, the triangle with sides [tex]5[/tex], [tex]12[/tex], [tex]13[/tex] has perimeter [tex]30[/tex] and area [tex]\frac{5\times 12}{2}=30[/tex]. Hence there are [tex]5[/tex] perfect triangles with integer sides.

Solution:
Let [tex]p^3-4p+9=n^2[/tex] with an integer [tex]n\ge 0[/tex]. Since [tex]p\ge 2[/tex], we have [tex]p^3-4p=p\left(p^2-4\right)\ge 0[/tex], so [tex]n\ge 3[/tex]. We rewrite the equation as
[tex]p(p-2)(p+2)=(n-3)(n+3)[/tex]
The prime [tex]p[/tex] divides [tex](n-3)(n+3)[/tex], so [tex]p\mid n-3[/tex] or [tex]p\mid n+3[/tex].
Case [tex]1[/tex]: [tex]n-3=kp[/tex], [tex]k\ge 0[/tex]. If [tex]k=0[/tex], then [tex]n=3[/tex] and [tex]p\left(p^2-4\right)=0[/tex], so [tex]p=2[/tex]. If [tex]k\ge 1[/tex], substituting [tex]n=kp+3[/tex] and dividing by [tex]p[/tex] gives
[tex]p^2-4=k(kp+6)\iff p\left(p-k^2\right)=6k+4\gt 0[/tex]
Then [tex]p\gt k^2[/tex], i.e. [tex]p-k^2\ge 1[/tex] and [tex]k^2\lt p\le p\left(p-k^2\right)=6k+4[/tex], which gives [tex]k\le 6[/tex]. Moreover, [tex]p[/tex] is a prime divisor of [tex]6k+4[/tex] greater than [tex]k^2[/tex]. For [tex]k=1[/tex] the prime divisors of [tex]10[/tex] are [tex]2[/tex] and [tex]5[/tex], but [tex]2\times (2-1)=2\ne 10[/tex] and [tex]5\times (5-1)=20\ne 10[/tex]. For [tex]k=2[/tex], [tex]4[/tex], [tex]5[/tex], [tex]6[/tex] no prime divisor of [tex]16[/tex], [tex]28[/tex], [tex]34[/tex], [tex]40[/tex] is greater than [tex]4[/tex], [tex]16[/tex], [tex]25[/tex], [tex]36[/tex], respectively. For [tex]k=3[/tex] we get [tex]p(p-9)=22[/tex], so [tex]p=11[/tex] (indeed, [tex]11\times 2=22[/tex]).
Case [tex]2[/tex]: [tex]n+3=kp[/tex], [tex]k\ge 1[/tex]. Substituting [tex]n=kp-3[/tex] and dividing by [tex]p[/tex] gives
[tex]p^2-4=k(kp-6)\iff p\left(k^2-p\right)=6k-4\gt 0[/tex]
So [tex]p[/tex] and [tex]q=k^2-p[/tex] are positive integers with [tex]p+q=k^2[/tex] and [tex]pq=6k-4[/tex]. From [tex](p-1)(q-1)\ge 0[/tex] we get [tex]p+q\le pq+1[/tex], i.e. [tex]k^2\le 6k-3[/tex], so [tex]k\le 5[/tex]. For [tex]k=1[/tex], [tex]2[/tex], [tex]4[/tex], [tex]5[/tex] there are no positive integers with sum [tex]1[/tex], [tex]4[/tex], [tex]16[/tex], [tex]25[/tex] and product [tex]2[/tex], [tex]8[/tex], [tex]20[/tex], [tex]26[/tex], respectively. For [tex]k=3[/tex] we have [tex]p+q=9[/tex], [tex]pq=14[/tex], so [tex]\{p, q\}=\{2, 7\}[/tex], i.e. [tex]p=2[/tex] or [tex]p=7[/tex].
Check:
[tex]2^3-8+9=9=3^2,\qquad 7^3-28+9=324=18^2,\qquad 11^3-44+9=1296=36^2[/tex]
Hence [tex]p\in\{2, 7, 11\}[/tex].

Solution:
If [tex]x\le -2[/tex], then [tex]0\lt 2^x+2^{2x+1}\le\frac{1}{4}+\frac{1}{8}\lt 1[/tex], so [tex]1\lt y^2\lt 2[/tex], which is impossible. For [tex]x=-1[/tex] we get [tex]y^2=2[/tex], also impossible. For [tex]x=0[/tex] we get [tex]y^2=4[/tex], [tex]y=\pm 2[/tex].
Let [tex]x\ge 1[/tex]. Since [tex](x, y)[/tex] and [tex](x, -y)[/tex] are solutions simultaneously, we may assume [tex]y\gt 0[/tex]. The left-hand side is odd, so [tex]y[/tex] is odd, and
[tex]2^x\left(1+2^{x+1}\right)=(y-1)(y+1)[/tex]
The even numbers [tex]y-1[/tex] and [tex]y+1[/tex] differ by [tex]2[/tex], so one of them is divisible by [tex]2[/tex] but not by [tex]4[/tex], and the other one is divisible by [tex]4[/tex]. Since [tex]1+2^{x+1}[/tex] is odd, the product contains the factor [tex]2[/tex] exactly [tex]x[/tex] times, so the other number equals [tex]2^{x-1}m[/tex] with [tex]m[/tex] odd, and [tex]x-1\ge 2[/tex], i.e. [tex]x\ge 3[/tex]. Thus [tex]y=2^{x-1}m+\varepsilon[/tex], where [tex]m[/tex] is an odd positive integer and [tex]\varepsilon=\pm 1[/tex]. Substituting into [tex]y^2-1=2^x+2^{2x+1}[/tex]:
[tex]2^{2x-2}m^2+2^x m\varepsilon=2^x+2^{2x+1}[/tex]
and dividing by [tex]2^x[/tex]:
[tex]2^{x-2}m^2+m\varepsilon=1+2^{x+1}\iff 2^{x-2}\left(m^2-8\right)=1-\varepsilon m[/tex]
If [tex]\varepsilon=1[/tex], the right-hand side is [tex]1-m\le 0[/tex], so [tex]m^2\le 8[/tex], hence [tex]m=1[/tex], and then [tex]-7\times 2^{x-2}=0[/tex], which is impossible.
If [tex]\varepsilon=-1[/tex], then [tex]2^{x-2}\left(m^2-8\right)=1+m\gt 0[/tex], so [tex]m\ge 3[/tex]. Since [tex]x\ge 3[/tex], we have [tex]2^{x-2}\ge 2[/tex] and
[tex]1+m\ge 2\left(m^2-8\right)\iff 2m^2-m-17\le 0\Rightarrow m\le 3[/tex]
So [tex]m=3[/tex], [tex]2^{x-2}=4[/tex], [tex]x=4[/tex] and [tex]y=2^3\times 3-1=23[/tex]. Check: [tex]1+2^4+2^9=1+16+512=529=23^2[/tex].
The solutions are [tex](0, 2)[/tex], [tex](0, -2)[/tex], [tex](4, 23)[/tex] and [tex](4, -23)[/tex].

Solution:
Suppose that natural numbers [tex]m[/tex], [tex]n[/tex] satisfy the equation. Then [tex]2^m\equiv 19\pmod{37}[/tex], and since [tex]2\times 19=38\equiv 1\pmod{37}[/tex], multiplying by [tex]2[/tex] we get
[tex]2^{m+1}\equiv 1\pmod{37}[/tex]
Let [tex]d[/tex] be the smallest positive integer with [tex]2^d\equiv 1\pmod{37}[/tex]. If [tex]2^N\equiv 1\pmod{37}[/tex] and [tex]N=qd+r[/tex] with [tex]0\le r\lt d[/tex], then [tex]2^r\equiv 1\pmod{37}[/tex], so [tex]r=0[/tex] by the minimality of [tex]d[/tex]; hence [tex]d[/tex] divides every such [tex]N[/tex], in particular [tex]d\mid m+1[/tex]. Moreover,
[tex]2^8=256\equiv -3\pmod{37},\qquad 2^{10}=1024\equiv 25\equiv -12\pmod{37}[/tex]
[tex]2^{18}=2^{10}\times 2^8\equiv (-12)(-3)=36\equiv -1\pmod{37},\qquad 2^{36}\equiv 1\pmod{37}[/tex]
Thus [tex]d\mid 36[/tex], but [tex]d\nmid 18[/tex] (otherwise [tex]2^{18}\equiv 1[/tex]). The divisors of [tex]36[/tex] that do not divide [tex]18[/tex] are [tex]4[/tex], [tex]12[/tex] and [tex]36[/tex], so [tex]4\mid d[/tex]. Then [tex]4\mid m+1[/tex], i.e. [tex]m=4j+3[/tex] with an integer [tex]j\ge 0[/tex], and
[tex]2^m=8\times 16^j[/tex]
ends in the digit [tex]8[/tex] (for [tex]j=0[/tex] it equals [tex]8[/tex], and for [tex]j\ge 1[/tex] the number [tex]16^j[/tex] ends in [tex]6[/tex] and [tex]8\times 6=48[/tex]). Hence [tex]37n^2=2^m-19[/tex] ends in the digit [tex]9[/tex]. Since [tex]37n^2\equiv 7n^2\pmod{10}[/tex] and [tex]7\times 3=21\equiv 1\pmod{10}[/tex], multiplying by [tex]3[/tex] we get
[tex]7n^2\equiv 9\pmod{10}\Rightarrow n^2\equiv 27\equiv 7\pmod{10}[/tex]
i.e. [tex]n^2[/tex] ends in the digit [tex]7[/tex]. But the square of an integer can end only in [tex]0[/tex], [tex]1[/tex], [tex]4[/tex], [tex]5[/tex], [tex]6[/tex] or [tex]9[/tex]. This contradiction proves that the equation has no solution in natural numbers.

Solution:
Suppose that integers [tex]x[/tex], [tex]y[/tex] satisfy the equation, and put [tex]d=x-y[/tex], [tex]s=x+y[/tex]. Then
[tex]x^5-y^5=dQ,\qquad Q=x^4+x^3y+x^2y^2+xy^3+y^4[/tex]
Substituting [tex]x=\frac{s+d}{2}[/tex], [tex]y=\frac{s-d}{2}[/tex], one checks the identity
[tex]16Q=5s^4+10s^2d^2+d^4[/tex]
Hence [tex]Q\ge 0[/tex], and from [tex]dQ=1993\gt 0[/tex] we get [tex]d\ge 1[/tex] and [tex]Q\ge\frac{d^4}{16}[/tex]. Therefore
[tex]1993=dQ\ge\frac{d^5}{16}\Rightarrow d^5\le 31888\lt 32768=8^5\Rightarrow d\le 7[/tex]
Also [tex]d[/tex] divides [tex]1993[/tex], which is odd and not divisible by [tex]3[/tex], [tex]5[/tex] or [tex]7[/tex]:
[tex]1993=3\times 664+1=5\times 398+3=7\times 284+5[/tex]
hence [tex]d=1[/tex]. For every integer [tex]t[/tex] we have [tex]t^5\equiv t\pmod 5[/tex]; it suffices to check the remainders [tex]0[/tex], [tex]1[/tex], [tex]2[/tex], [tex]3[/tex], [tex]4[/tex]:
[tex]0^5=0,\quad 1^5=1,\quad 2^5=32\equiv 2,\quad 3^5=243\equiv 3,\quad 4^5=1024\equiv 4\pmod 5[/tex]
Therefore [tex]x^5-y^5\equiv x-y=1\pmod 5[/tex], while [tex]1993\equiv 3\pmod 5[/tex]. This contradiction shows that the equation has no integer solutions.