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here are prefect squares:
1, 4, 9, 16, 25 ...

a perfect square is a whole number that can be expressed as another whole number multiplied by itself. for example,

25 = 5 x 5
35 = 6 x 6
49 = 7 x 7
...
...
100 = 10 x 10
121 = 11 x11

5 is not a perfect square. there is no whole number that when multiplied by itself gives 5. ancient greeks may have referred to this number as "rectangular". i find this cheeky and will call it a non-square.

here's the tough stuff.
can you find a perfect square that is twice the size of another perfect square?
can you find a perfect square that is thrice the size of another perfect square?
can you find a perfect square that is quadruple the size of another perfect square?
can you find a perfect square that is 169 times the size of another perfect square?

if you think any of these questions are impossible to answer, please elaborate
i don't know why i feel so dry
ugh im having a hard time answering the first question i cant seem to think of any

does that mean there are none or am i not getting it
banned
idk you tell me
i don't know why i feel so dry
Quote by Eastwinn
idk you tell me

man you are making my brain feel mushy again
banned

Quote by seventh_angel
169 and 28561. 28561 is 169 times the size of 169, which is 13x13. I don't want to think about the others.

are there others?
i don't know why i feel so dry
Quote by MinterMan22

heehehehe
banned
I don't come to UG to do math, Eastwinn.
This ends now, eat the goddamn beans!
math is just gay number sex
Quote by ManInTheBox14
The one that's quadruple another is 16.

are there others?
i don't know why i feel so dry
289 1156
324 1296
361 1444
400 1600
441 1764
484 1936
529 2116
576 2304
625 2500
676 2704

Some examples of 169 times:
1 169
4 676
9 1521
16 2704
25 4225
36 6084
49 8281

Double and triple shouldn't work since a product of perfect squares is always a perfect square, so the factors can never be non-perfect squares like 2 or 3.
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LET'S GO BUCKS
Last edited by AeroRocker at Jun 1, 2014,
okay easty i have come to the conclusion that there are no perfect squares twice the size as another perfect square because i have been thinking about it for a long time and i got nothing yo
banned
its hip to be square
okay easty i have come to the conclusion that there are no perfect squares twice the size as another perfect square because i have been thinking about it for a long time and i got nothing yo

yeah well if there aren't any why not?

aero: how many examples are there? if i give you a square, can you always quadruple it to get another square? you're explanation at the bottom of your post is a tad incomplete. i can tell you have the right idea though.
i don't know why i feel so dry
Quote by Eastwinn
yeah well if there aren't any why not?

i was hoping you would explain this actually
banned
Quote by Eastwinn

how many examples are there?

infinite

Quote by Eastwinn
if i give you a square, can you always quadruple it to get another square?

yep

If you want double or triple a perfect square then you are multiplying a perfect square by a non-perfect square which is always a non-perfect square.
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LET'S GO BUCKS
Last edited by AeroRocker at Jun 1, 2014,
Quote by AeroRocker
infinite

yep

Quote by AeroRocker
If you want double or triple a perfect square then you are multiplying a perfect square by a non-perfect square which is always a non-perfect square.

why?

i don't know why i feel so dry
Quote by AeroRocker

If you want double or triple a perfect square then you are multiplying a perfect square by a non-perfect square which is always a non-perfect square.

yeah what he said
banned
x^2 * y^2 = xy * xy

If x^2 is a perfect square and y^2 is not, then xy is not a whole number. amirite?
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LET'S GO BUCKS
okay, suppose x is whole, therefore x^2 is a whole number. y^2 is not a perfect square therefore y is not a whole number. all by definition. the result may be written in the form xy * xy but is there perhaps a whole r such r = xy * xy but r != xy ?

the more obvious it seems, the more "why?"s i'm gonna hand out.

edit: let me put this in the way i prefer:

given p and q are whole numbers, is there a solution to this equation:

2p^2 = q^2

substitute the 2 with whichever factor you want, but i think it helps to just start with two and go from there.
i don't know why i feel so dry
Last edited by Eastwinn at Jun 1, 2014,
for A*x^2=y^2, the only integer solution is zero. Since this can be solved as y=sqrt(A)*x, the solution will always be an irrational number unless A is a perfect square.
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LET'S GO BUCKS
Quote by AeroRocker
for A*x^2=y^2, the only integer solution is zero. Since this can be solved as y=sqrt(A)*x, the solution will always be an irrational number unless A is a perfect square.

why must it always be irrational if A is not a perfect square?
i don't know why i feel so dry
I wish I was in the right state of mind to participate in the thread...
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I wish I was in the right state of mind to participate in the thread...

it could go on for a while, we'll see. aerorocker may come back with the finisher or he might get stuck. meanwhile i still appreciate any and all confusion for when i write up my full explanation.
i don't know why i feel so dry
Every even perfect square is quadruple another perfect square, because 4 is itself a perfect square.
alright deads i'll give you an intuitive understanding of this problem.

suppose you have a perfect square N. its square root we will call r such that r x r = N.

we wish to find a perfect square M such that the following is true,
2 x N = M

this was the original question that you failed to find a pair N, M for. the answer comes as something called the fundamental theorem of arithmetic. here it is:

Quote by the fundy theorem
every whole number is equal to a UNIQUE product of prime numbers

examples:
6 = 3 x 2
12 = 2 x 2 x 3
42 = 2 x 3 x 7

these representations are unique (reordering them doesn't count) and they exist for every number. the representation of a prime, like 7, is just... 7.

since every number has a "prime factorization" and each prime factorization corresponds to only one number, they may be used interchangeably. i feel like this intuitive enough that i don't need to prove it. i will, if you're interested.

so what happens when you multiply a number by itself in terms of its prime factors? take 42 as an example.

42 x 42 = (2 x 3 x 7) x (2 x 3 x 7) = 2 x 3 x 7 x 2 x 3 x 7 = (2 x 2) x (3 x 3) x (7 x 7)

in that last bit, each number in parenthesis is a square. so you can see a useful property already, that a square number has an even number of factors. try this on paper a few times with perfect squares and with non-squares. soon enough it will be obvious. (you will also discover that having an even number of factors is not enough to make you a square!)

so back to 2N = M
because r x r = N we can write 2 x r x r = M. in order to show that M can be a perfect square, we need to rewrite it in terms of, say, s x s = M.

that is, we need to find an s such that 2 x r x r = s x s.

if two numbers are equal, by the fundamental thereon, they must have the same prime factors. s x s has an even number of prime factors, as we discovered above, and for the same reason r x r has an even number of prime factors as well BUT 2 x r x r has an extra factor of 2 which means it has an odd number of prime factors. thus, the prime factorization of 2 x r x r MUST be different than s x s so there is no solution to 2N = M in whole numbers. that's it, the 2 is a wrench thrown in that cannot be absorbed.

a slightly different argument must be used for other number besides 2 but you should be able to work it out yourself. as a corollary, we can now prove that the square root of two is irrational.

first, ASSUME that it is rational, then by definition sqrt(2) = p/q where p and q are whole numbers. we can limit our search for the positive square root of 2 and let p and q both be positive, just for simplicity's sake. now take that equation

sqrt(2) = p/q

and square both sides. the square root of two squared is by definition two. thus,

2 = p/q x p/q = (p x p)/(q x q)

to get rid of this fraction nonsense, multiply both sides by (q x q) ..

2(q x q) = (p x p)

if we replace (q x q) with a perfect square N = q x q and likewise a perfect square M = (p x p) we have

2N = M

the exact equation we just proved had no solution. thus there cannot be a p/q where p and q are whole that equals the square root of two. this can be extended further if you extend the above theorem. you can even generalize it to cube roots, quartics, and so on.

unfortunately the irrationality of pi is much more complicated. have a nice day
i don't know why i feel so dry
ITT: Nerds

Actually, I watch Numberphile on Youtube a lot and really enjoy it. But ^ proofs are lame.
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ITT: Nerds

Actually, I watch Numberphile on Youtube a lot and really enjoy it. But ^ proofs are lame.

they're the only good part about math
i don't know why i feel so dry

i found the perfect square
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Hello.
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i found the perfect square

also a cube but ya know
i don't know why i feel so dry
Ts, you need to learn how to equations.
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This thread's ballsack needs to get slapped
My God, it's full of stars!