Doubt 1- in Diophantine Equation

Solve the equation $x^{2}+1=y^{3}$ in positive integers.

note: I tried this for a few hours but the answer didn't come. Please clarify. Thanks!

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Math	Appears as
Remember to wrap math in `$` ... `$` or `\[` ... `\]` to ensure proper formatting.
`2 \times 3`	$2 \times 3$
`2^{34}`	$2^{34}$
`a_{i-1}`	$a_{i-1}$
`\frac{2}{3}`	$\frac{2}{3}$
`\sqrt{2}`	$\sqrt{2}$
`\sum_{i=1}^3`	$\sum_{i=1}^3$
`\sin \theta`	$\sin \theta$
`\boxed{123}`	$\boxed{123}$

Comments

There are no solutions.

Lets look at the remainder if you devide by 4 (modular arithmetic). A square number has always a remainder of 0 or 1 if you devide by 4, therefore the left hand side has always a remainder of 1 or 2 if you devide by 4. A cubed number has always a remainder of 0 or 1 or 3 if you devide by 4. If x and y are solutions, then the left hand side and the right hand side must have the same remainder if you devide by 4. Hence they must have a remainder of 1. And hence ${ x }^{ 2 }\equiv 0$ if you devide by 4 (because ${ x }^{ 2 }+1\equiv 1$ ).

If ${ x }^{ 2 }\equiv 0$ then $x$ must have a remainder of 0 or 2. Remember that fact

Because we are talking about positive integers we can deduce that $y>x$ fro all solutions. And hence there must be a positive integer $c$ for which $y=x+c$ holds true. ${ x }^{ 2 }+1={ \left( x+c \right) }^{ 3 }={ x }^{ 3 }+3{ x }^{ 2 }c+3x{ c }^{ 2 }+{ c }^{ 3 }$

Now there are two cases: $x$ has a remainder of 0 or 2.

CASE 1: $x$ has a remainder of 0 if you devide by 4

We can say that ${ x }^{ 2 }+1\equiv 1$ and ${ x }^{ 3 }+3{ x }^{ 2 }c+3x{ c }^{ 2 }+{ c }^{ 3 }\equiv { c }^{ 3 }$ and hence: $1={ c }^{ 3 }$ Now we know in case one $c$ must be 1. So ${ x }^{ 2 }+1={ x }^{ 3 }+3{ x }^{ 2 }c+3x{ c }^{ 2 }+{ c }^{ 3 }$ simplifies to ${ x }^{ 2 }={ x }^{ 3 }+3{ x }^{ 2 }+3x$ which has no posivite integer roots for x.

NO SOLUTIONS IN CASE ONE

CASE 2: $x$ has a remainder of 2 if you devide by 4

We can say that ${ x }^{ 2 }+1\equiv 5\equiv 1$ and ${ x }^{ 3 }+3{ x }^{ 2 }c+3x{ c }^{ 2 }+{ c }^{ 3 }\equiv 8+12c+6{ c }^{ 2 }+{ c }^{ 3 }\equiv 2{ c }^{ 2 }+{ c }^{ 3 }$ Hence $1=2{ c }^{ 2 }+{ c }^{ 3 }$ But this has no positive integer roots for $c$ and hence

NO SOLUTION IN CASE TWO

ANd because there are no solutions in all cases, there are no solutions in general. I'm not 100% sure wether this is correct, but if you have any questions I try to answer it the best I can ;)

CodeCrafter 1 - 1 year, 3 months ago

This is 120% correct. I understood it. Good job

Nitin Kumar - 1 year, 3 months ago

$x^2 + 1 = y^3$

$x = 0, y = 1$

$0^2 + 1 = 1^3$

$0 + 1 = 1$

$1 = 1$

This is my solution.

A Former Brilliant Member - 1 year, 1 month ago

The problem asked for positive solutions. 0 is not positive.

Otherwise it would be too easy ;-)

CodeCrafter 1 - 1 year, 1 month ago

Math	Appears as
Remember to wrap math in `\(` ... `\)` or `\[` ... `\]` to ensure proper formatting.
`2 \times 3`	$2 \times 3$
`2^{34}`	$2^{34}$
`a_{i-1}`	$a_{i-1}$
`\frac{2}{3}`	$\frac{2}{3}$
`\sqrt{2}`	$\sqrt{2}$
`\sum_{i=1}^3`	$\sum_{i=1}^3$
`\sin \theta`	$\sin \theta$
`\boxed{123}`	$\boxed{123}$