Integral solutions

Here are a few interesting problems :

(1)(1) Prove that there are infinitely many triples (x,y,z)∈Z+(x,y,z) \in \mathbb{Z^{+}} such that

x3+y4=z7x^3+y^4=z^7

(2)(2) Show that there are no solutions (a,b,c)∈Z(a,b,c) \in \mathbb{Z} such that

a2+b2−8c=6a^2+b^2-8c=6

(3)(3) Determine all triples (x,y,z)∈Z+(x,y,z) \in \mathbb{Z^{+}} satisfying

2x2y2+2y2z2+2z2x2−x4−y4−z4=5762x^2y^2+2y^2z^2+2z^2x^2-x^4-y^4-z^4 = 576

#NumberTheory

Note by Ankit Kumar Jain
4 years, 1 month ago

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Comments

(1)(1) :

Set x=24a,y=23a,z=2bx=2^{4a},y=2^{3a},z=2^{b} , which leads on simplification to the linear diophantine equation 7b−12a=17b-12a=1 which has infinitely many solutions since (12,7)=1(12,7)=1 . One such being (x,y,z)=(216,212,27)(x,y,z)=(2^{16},2^{12},2^{7})

(2)(2) :

Given, a2+b2=8c+6  ≡  6  (mod  8)\displaystyle a^2+b^2=8c+6\;\equiv\; 6\;(mod \; 8) . Now it is evident that for any integer nn the relation,

n2  ≡  0,1,4  (mod  8)\displaystyle n^2\;\equiv\;0,1,4\;(mod\; 8) is satisfied which clearly contradicts a2+b2  ≡  6  (mod  8)a^2+b^2\;\equiv\;6\;(mod \;8) . Thus no solutions.

Aditya Narayan Sharma - 4 years, 1 month ago

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Good ones.

Ankit Kumar Jain - 4 years, 1 month ago

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You can try out my note Real Roots.

Ankit Kumar Jain - 4 years, 1 month ago

Do post your solutions guys!!

Ankit Kumar Jain - 4 years, 1 month ago

(3):(3) :

(x+y+z)(x+y−z)(y+z−x)(z+x−y)=2x2y2+2y2z2+2z2x2−x4−y4−z4  ⟹  (x+y+z)(x+y−z)(y+z−x)(z+x−y)=576\begin{aligned}(x+y+z)(x+y-z)(y+z-x)(z+x-y)&=2x^2y^2+2y^2z^2+2z^2x^2-x^4-y^4-z^4\\ \implies (x+y+z)(x+y-z)(y+z-x)(z+x-y)&=576\\\\ \end{aligned}

Let, x+y+z=a\text{Let, } x+y+z=a\\

a(a−2x)(a−2y)(a−2z)=576(1)  ⟹  All the factors are even(2)Since a,a-2x,a-2y,a-2z are all of the same paritya−2x+a−2y+a−2z=3a−2(x+y+z)=3a−2a=a  ⟹  The largest factor is the sum of the other 3 factors(3)\begin{aligned}a(a-2x)(a-2y)(a-2z)&=576 \hspace{7mm}\color{#3D99F6}\small (1)\\\\ \implies \text{All the factors are }\text{even}&\hspace{7mm}\color{#3D99F6}\small (2)\hspace{7mm}\text{Since a,a-2x,a-2y,a-2z are all of the same parity}\\\\ a-2x+a-2y+a-2z&=3a-2(x+y+z)\\\\ &=3a-2a=a\end{aligned}\\\\ \implies \text{The largest factor is the sum of the other 3 factors}\hspace{7mm}\color{#3D99F6}\small (3)

The only unordered pair that satisfy(1),(2) and (3)is,(12,6,4,2)  ⟹  (x,y,z)can take the values (3,4,5),(3,5,4),(4,3,5),(4,5,3),(5,3,4),(5,4,3)\begin{aligned}\text{The only unordered pair that satisfy}\small \color{#3D99F6}(1),(2) \normalsize\color{#333333}\text{ and }\small\color{#3D99F6}(3) \color{#333333}\normalsize \text{is,}(12,6,4,2)\\ \implies (x,y,z) \text{can take the values } (3,4,5),(3,5,4),(4,3,5),(4,5,3),(5,3,4),(5,4,3)\end{aligned}

Anirudh Sreekumar - 4 years, 1 month ago

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Nice solution.

Ankit Kumar Jain - 4 years, 1 month ago
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