Question

how many solution of n (integer positive) so that n^2 + (n+1)^2 be a perfect square, with n < 200

Answer 1

@rational

Answer 2

i just remember the triplet of pythagoras which is consecutive number :
(3,4,5), (20,21,29). any else ?

Answer 3

and i need a way to get n

Answer 4

hmmm I'm only going to think here
\[n^2+(n+1)^2=k^2 \\ n^2+n^2+2n+1=k^2 \\ 2n^2+2n+1-k^2 =0 \\ n=\frac{ -2 \pm \sqrt{4-4(2)(1-k^2)}}{2(2)} \\ n=\frac{-2 \pm \sqrt{4} \sqrt{1-2(1-k^2)}}{4} \\ n=\frac{-1 \pm \sqrt{1-2+2k^2}}{2} \\ n=\frac{-1 \pm \sqrt{2k^2-1}}{2}\]
so far I don't know if this helps right now
this is just the first thing I thought of

Answer 5

we want to choose integer k so that n is also an integer

Answer 6

we need 2k^2-1 to be a perfect square

Answer 7

also 2k^2-1 is odd

Answer 8

so we know for some odd integer m
we have
m^2=2k^2-1
right?

Answer 9

im thinking of first generating all pythagorean triples \((a,b,c)\) satisfying given conditions with even \(n\) :
\[a=n = 2xy\\b=n+1 = x^2-y^2 = 2xy+1\\c=x^2+y^2\]

Answer 10

you mean n = a^2 - b^2, (n+1) = 2ab, and (a^2 + b^2) right ?

Answer 11

i am choosing "n" to be even
it doesn't matter

Answer 12

if n = a^2 - b^2
n + 1 = 2ab
-------------- (-)
-1 = a^2 - 2ab - b^2
a^2 - 2ab - b^2 + 1 = 0
a^2 + b^2 - 2ab - 2b^2 + 1 = 0
(a - b)^2 = 2b^2 - 1
a - b = +- sqrt (2b^2 - 1)
a = b +- sqrt(2b^2 - 1)
hmmm, what's next ?

Answer 13

Fine that works too

Answer 14

I feel like I hit a brick wall with my thingy up above

Answer 15

this is a hard problem haha

Answer 16

I mean I can choose integer k so 2k^2-1 is an odd perfect square
and hopefully the numerator I had is divisible by 2
but that sounds like a lot of numbers to plugin between 1 and 200

Answer 17

ok, thanks @freckles . so far i got k = 1 or k = 5

Answer 18

i want to conecting it with my job above :
a = b +- sqrt(2b^2 - 1)
for b = 2 than a = 1 + 1 = 2, so (a,b) = (2,1) give us n = 2^2 - 1^2 = 3
for b = 5 than a = 5 + 7 = 12, so (a,b) = (5,12) get n = 119

Answer 19

so far i got n = 3,20,19

Answer 20

but not sure if any else for solution of n

Answer 21

can i replace the general of triples : a^2 - b^2 , 2ab, a^2 + b^2 be
2ab, a^2 - b^2, a^2 + b^2 ? @rational

Answer 22

i mean in order n = 2ab and (n+1) = a^2 - b^2

Answer 23

and obvious a^2 + b^2 is always be the longest side (hypotenus), so i just can play in order 2ab and a^2 - b^2 only, right ?

Answer 24

let me try this :)

Answer 25

n = 2ab
n + 1 = a^2 - b^2
----------------- (-)
-1 = 2ab - a^2 + b^2
a^2 - 2ab - b^2 = 1
a^2 - 2ab + b^2 = 2b^2 + 1
(a - b)^2 = 2b^2 + 1
a = b +- sqrt(2b^2 + 1)
a = b +- sqrt(2b^2 + 1)

Answer 26

ah, yes b = 2 is work

Answer 27

http://www.wolframalpha.com/input/?i=solve+n%5E2+%2B+%28n%2B1%29%5E2+%3D+k%5E2%2C+1%3Cn%3C200+over+integers

Answer 28

so, (a,b) = (5,2)
get n = 2ab = 2(5)(2) = 20

Answer 29

so, only 3 solutions satisfied for n = 3,20,119
case closed :)

Answer 30

wolfram is make sure to me :)
thanks a lot @rational

Answer 31

I did them all through brute force; yes, there are only 3 solutions.

Answer 32

i dont know what's brute force method but thanks to make sure too :)

Answer 33

there is a nice way to work this if you know infinite continued fractions of irrational numbers

Answer 34

\(n^2+(n+1)^2=k^2 \\ 2n^2+2n+1=k^2\\4n^2+4n+2 = 2k^2\\(2n+1)^2-2k^2 = -1\)

This is pell's equation, let \(2n+1=x\) :
\[x^2-2k^2=-1\]

Answer 35

Brute force is starting at 1 and doing every one until you hit 200.

Answer 36

All the solutions to above equation are given by \(\large (p_{2i}, ~q_{2i})\) where \(\dfrac{p_{i}}{q_i}\) is the \(i\)th sum of continued continued fraction of \(\sqrt{2}\) :
\[\dfrac{p_i}{q_i} = 1+\dfrac{1}{2+\dfrac{1}{2+\dfrac{1}{2+\cdots \text{(i times)} }}}\]

Answer 37

the first solution to \(x^2 - 2k^2 = -1\) can be obtained working \(2\)nd sum :
\[\dfrac{p_2}{q_2} = 1+\dfrac{1}{2+\dfrac{1}{2}} = \dfrac{7}{5}\]

therefore \((7,5)\) is a solution to \(x^2 - 2k^2 = -1\)
\(x=2n+1 = 7 \implies n = 3\)

similarly working \(\dfrac{p_4}{q_4}\) gives the next solution