Question

Prove/Disapprove

Answer 1

\(\large \begin{align} \color{black}{\normalsize \text{Prove\Disapprove that the equation has no integer solution } \hspace{.33em}\\~\\
x^2-y^2=2(y_1 \times y_2 \times y_3 \times y_4 \times \cdot \cdot \cdot \cdot \times y_n) \hspace{.33em}\\~\\
\normalsize \text{ where } \hspace{.33em}\\~\\
\{y_1 , y_2 , y_3 , y_4 , \cdot \cdot \cdot \cdot , y_n\} \hspace{.33em}\\~\\
\normalsize \text{ are all odd natural numbers } \hspace{.33em}\\~\\
}\end{align}\)

Answer 2

^_^

Answer 3

\[\Huge \begin{array}l\color{red}{\text{^}}\color{orange}{\text{-}}\color{#E6E600}{\text{^}}\color{green}{\text{}}\end{array} \]

Answer 4

I will give a counter example or maybe I got the question wrong
x=5,y=4
x^2-y^2=9 .
9 isn't divisible by 9.

Answer 5

I meant by 2

Answer 6

but 2 isn't the multiple of 9

Answer 7

So the equation is wrong !

Answer 8

i think u misinterpreted the question may be

Answer 9

I think so.I can't be that easy :)
Can you make it more clear?

Answer 10

ok

Answer 11

for example prove that the equation has no solution
\(\large \begin{align} \color{black}{x^2-y^2=2\times 3\times 5 \hspace{.33em}\\~\\
}\end{align}\)

here \(3\) and \(5\) are odd

Answer 12

is it clear ?

Answer 13

Yea

Answer 14

\[x^2 -y^2 = (x+y)(x-y)\]
^_^ :D

Answer 15

Let's try proving it:
odd*odd = odd.
To get even when subtracting
x is even and y is even
x is odd and y is odd
\[x=\sqrt{2*odd+y ^{2}}\]
now let's say y is an even number
y=2p
\[x=\sqrt{2(odd+p*2p}=\sqrt{2(odd+even}=\sqrt{2*odd} = \sqrt{2}*\sqrt{odd}\]
Thus it is proved for y is an even

y is an odd:
\[x^2-y^2= 2*odd.\]
add zero -1+1
then divide by 8
\[x^2-1-(y^2-1)=2* odd\]
\[\frac{ x^2-1-(y^2-1) }{ 8 }=\frac{2* odd }{ 8 }.\]
left side is an integer while the left isn't,so this proves the equation :D

Answer 16

@mathmath333

Answer 17

for the last line do u meant the
left side is an integer while the right side isn't ?

Answer 18

ur first part is flawless for y is even.

Answer 19

but how does it proves that


\(\large \begin{align} \color{black}{\frac{ x^2-1-(y^2-1) }{ 8 }\hspace{.33em}\\~\\}\end{align}\)

is an integer.

Answer 20

Quick proof:

As above, nove x^2 - y^2 = (x+y)(x-y)

If one of x and y is even, and the other odd, both terms in this product are odd, so their product is odd - contradiction.

If they are both even (or odd), both terms in this product are eve, so their product is a multiple of 4 - contradiction.

QED

Answer 21

Haha I just noticed he said Prove/Disapprove. I will take the second option and simply disapprove of this math question.

Answer 22

@mathmath333
any odd square -1 is divisible by 8

Answer 23

ok u assumed \(y\) is odd so \(\dfrac{y^2-1}{8}\) is integer.

but how can \(\dfrac{x^2-1}{8}\) will also be an integer. which is attached with it.

Answer 24

we have x^2-y^2=2*odd= even
that implies that both x and y are even or both are odd.
and square of even is even and square of odd is odd.
http://www.mathsisfun.com/numbers/even-odd.html