Yet another integral problem

Prove that there are infinitely many perfect squares of the form

1+2x2+2y21+2^{x^{2}}+2^{y^{2}}

where (x,y)∈Z(x,y) \in \mathbb{Z}

#NumberTheory

Note by Ankit Kumar Jain
4 years, 1 month ago

No vote yet
1 vote

  Easy Math Editor

This discussion board is a place to discuss our Daily Challenges and the math and science related to those challenges. Explanations are more than just a solution — they should explain the steps and thinking strategies that you used to obtain the solution. Comments should further the discussion of math and science.

When posting on Brilliant:

  • Use the emojis to react to an explanation, whether you're congratulating a job well done , or just really confused .
  • Ask specific questions about the challenge or the steps in somebody's explanation. Well-posed questions can add a lot to the discussion, but posting "I don't understand!" doesn't help anyone.
  • Try to contribute something new to the discussion, whether it is an extension, generalization or other idea related to the challenge.
  • Stay on topic — we're all here to learn more about math and science, not to hear about your favorite get-rich-quick scheme or current world events.

MarkdownAppears as
*italics* or _italics_ italics
**bold** or __bold__ bold

- bulleted
- list
  • bulleted
  • list

1. numbered
2. list
  1. numbered
  2. list
Note: you must add a full line of space before and after lists for them to show up correctly
paragraph 1

paragraph 2

paragraph 1

paragraph 2

[example link](https://brilliant.org)example link
> This is a quote
This is a quote
    # I indented these lines
    # 4 spaces, and now they show
    # up as a code block.

    print "hello world"
# I indented these lines
# 4 spaces, and now they show
# up as a code block.

print "hello world"
MathAppears as
Remember to wrap math in \( ... \) or \[ ... \] to ensure proper formatting.
2 \times 3 2×3 2 \times 3
2^{34} 234 2^{34}
a_{i-1} ai−1 a_{i-1}
\frac{2}{3} 23 \frac{2}{3}
\sqrt{2} 2 \sqrt{2}
\sum_{i=1}^3 ∑i=13 \sum_{i=1}^3
\sin \theta sin⁡θ \sin \theta
\boxed{123} 123 \boxed{123}

Comments

1+2x2+2y2=1+2⋅2x2−1+2y2now let, y2=2⋅(x2−1)  ⟹  1+2x2+2y2=1+2⋅2x2−1+22⋅(x2−1)=1+2⋅2x2−1+(2(x2−1))2=(1+2(x2−1))2\begin{aligned}1+2^{x^2}+2^{y^2}&=1+2\cdot 2^{x^2-1}+2^{y^2}\\\\ \text{now let, }y^2&=2\cdot (x^2-1)\\\\ \implies 1+2^{x^2}+2^{y^2}&=1+2\cdot 2^{x^2-1}+2^{2\cdot (x^2-1)}\\\\ &=1+2\cdot 2^{x^2-1}+\left(2^{(x^2-1)}\right)^2\\\\ &=\left(1+2^{(x^2-1)}\right)^2\end{aligned}

Thus the above equation is always a perfect square when,

y2=2⋅(x2−1)y^2=2\cdot (x^2-1)

Rewriting we get,

x2−y22=1\begin{aligned}x^2-\dfrac{y^2}{2}=1\end{aligned}

This is similar to the pell equation which is of the form x2−ny2=1x^2-ny^2=1, where nn is an integer

if we make the substitution y22=2z2\dfrac{y^2}{2}=2 z^2, we can reduce it to the form of a pell equation in x,zx,z

now we have x2−2z2=1x^2-2z^2=1

we can find the fundamental solution of the pell equation by writing the sequence of convergents (hiki\dfrac{h_{i}}{k_{i}}) to the continued fractions of n\sqrt{n} .If (x1,z1)(x_{1},z_{1}) is the fundamental solution of the pell equation then x1=hix_{1}=h_{i} and z1=kiz_{1}=k_{i} for some ii such that xx is minimal

2=1+12+12+12+12+12+⋱\sqrt 2=1+\dfrac{1}{2+\dfrac{1}{2+\dfrac{1}{2+\dfrac{1}{2+\dfrac{1}{2+_\ddots}}}}}

we have h1k1=32\dfrac{h_1}{k_1}=\dfrac{3}{2}

(32−2⋅22)=1(3^2-2\cdot2^2)=1

we find that (3,2)(3,2) is the fundamental solution to the given pell equation

Once the fundamental solution is found the general solution (xk,zk)(x_{k},z_{k}) can be calculated by the relation ,

(xk+zkn)=(x1+z1n)k(x_{k}+z_{k}\sqrt n)=(x_{1}+z_{1}\sqrt n)^k

in the above case we get,

(xk+zk2)=(3+22)k(x_{k}+z_{k}\sqrt 2)=(3+2\sqrt 2)^k

similarly we have,

(xk−zk2)=(3−22)k(x_{k}-z_{k}\sqrt 2)=(3-2\sqrt 2)^k,as (xk+zk2)×(xk−zk2)=1(x_{k}+z_{k}\sqrt 2)\times(x_{k}-z_{k}\sqrt 2)=1

thus we get,

xk=(3+22)k+(3−22)k2zk=(3+22)k−(3−22)k22 x_{k}=\dfrac{(3+2\sqrt 2)^k+(3-2\sqrt 2)^k}{2}\\\\ z_{k}=\dfrac{(3+2\sqrt 2)^k-(3-2\sqrt 2)^k}{2\sqrt 2}

since we assumed ,y22=2z2\dfrac{y^2}{2}=2 z^2

we have y=2zy=2z

  ⟹  yk=2zk=(3+22)k−(3−22)k2\implies y_{k}=2 z_{k}=\dfrac{(3+2\sqrt 2)^k-(3-2\sqrt 2)^k}{\sqrt 2}

Thus for all numbers of the form

(xk,yk)=((3+22)k+(3−22)k2,(3+22)k−(3−22)k2)( x_{k},y_{k})=\left(\dfrac{(3+2\sqrt 2)^k+(3-2\sqrt 2)^k}{2},\dfrac{(3+2\sqrt 2)^k-(3-2\sqrt 2)^k}{\sqrt 2}\right)

1+2x2+2y21+2^{x^2}+2^{y^2} will be a perfect square

Anirudh Sreekumar - 4 years, 1 month ago

Log in to reply

That is complete brilliance..Mind blowing solution...Just phenomenal..

Ankit Kumar Jain - 4 years, 1 month ago

Post your solution guys

Ankit Kumar Jain - 4 years, 1 month ago
×

Problem Loading...

Note Loading...

Set Loading...