Question

Prove that if n is a perfect square, then n +2 is not a perfect square

Answer 1

i suppose first step would be to assume n is perfect square

Answer 2

so n = m^2

Answer 3

then i suppose i find n^2

Answer 4

i mean n + 2

Answer 5

n + 2 would be m^2 + 2

Answer 6

so now... i assume m^2 + 2 is a perfect square...

Answer 7

we have no 2 perfect square numbers with difference 2

Answer 8

so i let m^2 + 2 = k^2

Answer 9

@mukushla that's not really a proof.....

Answer 10

you may continue @igba

Answer 11

hmm i suppose the next step would be

m^2 - k^2 = 2

so (m+k)(m-k) = 2

i don't think this proves anything...

Answer 12

actually that is :)
\[m^2-k^2=2\]\[(m-k)(m+k)=2\]and this is impossible

Answer 13

why so?

Answer 14

well,,you've almost proved it..

now m and k both are integers/.

Answer 15

and its k^2 -m^2 = 2 and not otherwise..

Answer 16

oh yes

Answer 17

2 = (k + m)(k-m)

Answer 18

now...integers huh...

Answer 19

\[m+k>m-k\]so\[m+k=2\]so\[m=k=1\]but it gives\[m-k=0\]

Answer 20

you know k>m..

so from
(k+m)(k-m)= 1*2

=>k+m =2
and
k-m = 1

this doesnt have an integral solution..

Answer 21

why do i know k > m again?

Answer 22

m^2 + 2 = k^2

both m and k are positive integers

hence k>m

Answer 23

in layman language,, k^2 - m^2 >0

Answer 24

when was it assumed that m and k are positive?

Answer 25

what else do you mean by perfect square ?

Answer 26

i don't see how that relates.... the definition of a perfect square is x = k^2

Answer 27

4 is a perfect sq since its 2^2 i.e. square of a positive integer..
and likewise..

Answer 28

@mukushla
"we have no 2 perfect square numbers with difference 2"

Because 0,1,4,9,6,5,6,9,4,1 etc....

Answer 29

(-2)^2 is also 4...

Answer 30

you take the absolute value

Answer 31

i mean you have to..

Answer 32

why?

Answer 33

see,,
(k+m)(k-m) =2

either both are positive are both are negative,,since RHS is postive..

so even if you take -ve values,, negative * negative = positive ,,i.e. treated as absolute value..

Answer 34

...i don't think this is a valid proof....

Answer 35

another explanation might me x is a perfect square because its y^2

x = y^2 = (-y)^2

sqrt(x) = | y | = | -y |

Answer 36

who can writing here now two perfect squares with difference of 2 ?

so because i think do not exist or .... ???

Answer 37

since when did sqrt x become |-y|

Answer 38

0 to 9 squares end in 0,1,4,9,6,5,6,9,4,1 and pattern repeats.

No +2 in there.

Answer 39

so and i think that this difference of two will be the ,,key" of this proof of ...

Answer 40

@lgbasallote what is your opinion from this ,please ?

Answer 41

my opinion is that this is some tricky algebra

Answer 42

how do you think it ???

Answer 43

so and why ?

Answer 44

because i can't think of a proof

Answer 45

how can you prove/disprove that the difference of two squares is 2...

Answer 46

hmmm...wait...

Answer 47

so i think this is very easy logicaly but to prove it will be difficile i think

so but there are again @saifoo.khan and @satellite73

probably they will can doing it

hope so much

Answer 48

their fields of specialty aren't exactly in proving

Answer 49

What's wrong with my proof?

Answer 50

anyway...

m^2 + 2 = k^2

k^2 - m^2 = 2

for it to be equal to 2...k^2 and m^2 should either be both even...or both odd

so if i assume both are even..

then k and m are also both even

so i can rewrite this as

(2x)^2 - (2y^2) = 2

4x^2 - 4y^2 = 2

then...

2x^2 - 2y^2 = 1

x^2 - y^2 = 1/2

then since it's x^2 and y^2 are both integers...the difference should also be integer

\('\therefore\) contradiction

Answer 51

and i believe your proof @jhonyy9 did not involve proving...but brute force substitution..

Answer 52

@estudier i mean

Answer 53

@mukushla too

Answer 54

So, that doesn't mean it's invalid..

Answer 55

it does actually.....you can't do infinite substitutions to prove anything....

Answer 56

There is no infinity involved, all integer squares end in the numbers I gave.

Answer 57

infinite loops involves infinity agree?

Answer 58

No infinite loop either, just logic.

Answer 59

An integer ends in one of digits 0 to 9.

The squares must end in etc....

Answer 60

but we both know that setting values is illegal in proving

Answer 61

I am not setting any values.

Answer 62

and actually... you were checking 1^2, 2^2, 3^2, 4^2, etc. thus that will go on forever

Answer 63

you're checking the squares of each number

Answer 64

I am not.

Answer 65

and though there is a pattern...there is still susbtitution involved

Answer 66

and then you'll have to substitute forever to verify that the pattern really does not break

Answer 67

then infinity is involved

Answer 68

It's not a pattern, it's a fact

0 to 9 squares end in 0,1,4,9,6,5,6,9,4,1

Answer 69

so you do admit that you checked the 0 to 9 squares

Answer 70

Anything else is impossible.

Answer 71

...that's substitution

Answer 72

anyway...my point is...substitution isn't really allowed in proving

Answer 73

You will find that this fact is used a lot in many valid proofs in number theory.

Answer 74

perhaps

Answer 75

OK, I can also do the algebra proof, but that doesn't mean it's any better than mine.

Answer 76

since algebra is applicable in all numbers set to it....wouldn't that make it better than brute force?

Answer 77

I am not using brute force, brute force is something like testing or experimental math.

Or exhaustion (which is also a valid proof technique).

It's the use of a fact about squares (ALL squares)

Answer 78

can you prove it? that it's a fact about squares?

Answer 79

Yes, all numbers end in 0 to 9 and so their squares end in .....QED.

Answer 80

you didn't really demonstrate anything...you just repeated what you said

Answer 81

You asked for proof, I just gave u a proof.

Answer 82

If you deny my proof, provide a counterexample.

Answer 83

since you try so hard to defend your proof, i'll accept it

Answer 84

That's very gracious, thank you:-)

Answer 85

do check on my algebra proof though please. i would like to know if that's the way contradiction works

Answer 86

OK, I will look at it now....

Answer 87

Did u do the case when both are odd?

(Perhaps a more straightforward way is one you start with n= some k squared then

next up must be (k+1)^2 = k^2 + 2k +1 and see what you can deduce from that...)

Answer 88

why do i have to do odd too? if i already disproved using even...doesn't that prove by contradiction already?

Answer 89

No because you divided the problem in cases, means you have to prove both (else you are missing all those cases)

Answer 90

oh i see. but that kind of proof is right? because i'm a little skeptic if stating "2 can only be an outcome of even and even or odd and odd" is a valid statement

Answer 91

again i ask... how is k > m

Answer 92

(k+1)^2 = k^2 + 2k +1 >= n +3 > n+2 QED

Answer 93

factoring involves lesser steps...its obvious that k>m
\[k^2−m^2=2\]\[(k−m)(k+m)=2\]easily u can see
\[k-m=1\]\[k+m=2\]add them\[2k=3\]
Contradiction

Answer 94

again i ask for the nth time @mukushla how is k > m

Answer 95

ok man i just want to answer ur question

Answer 96

"2 can only be an outcome of even and even or odd and odd" is a valid statement

Why do you think it might not be valid?

Answer 97

well because 8 - 4 is not 2 but it's even minus even