Sums of squares of two consecutive integers.......

.E_C_K_S.

Hi guys,
Just having a little trouble and was wondering can someone point me in the right direction.( I know that asking for solutions is not allowed)

I have to show that the sum of the squares of two consecutive integers can be a square and to find all examples.One example would be 3^2+4^2=5^2.etc

I have let my integers be k and k+1 squared both and have shown that the result is an odd number. Not really sure how to show it is a square or to find all examples. Would there be some formula I have overlooked here?

I will have to generalise this to n consecutive integers and higher powers eventually but if I get a good grounding for the sum of the squares first, I should be able to apply this later on.

Thanks!

Eliot Rosewater

Des McHale said not to use the Internet young man!

.E_C_K_S.

Hmmmm I wonder. I just need some guidance that is all!

Eliot Rosewater

It was only a joke! :pac:


As regards guidance, I've taken out a few books on number theory from the library. Heaving going though.

.E_C_K_S.

But k^2 +(k+1)^2 is not equal to (k+2)^2? Stuck now again!

Fremen

I'd start by looking at the argument that constructs all the Pythagorean triples, and try to modify it.

Sorry about my earlier reply, I misunderstood what you were trying to do.

.E_C_K_S.

Cheers, Ya I have started to look the construction of pythagoran triples ie (3,4,5) in general and will try find out more about the consecutive cases.

Fremen

Basic questions about Pythagorean triples can actually take you reasonably deep into number theory, as far as I know.

This is a fairly friendly introduction that skips over the abstractions and technicalitles and gets right to the point:
http://mathforum.org/dr/math/faq/faq.pythag.triples.html

.E_C_K_S.

I found a method to find the square of the sum of two consecutive integers i.e. like 3^2 +4^2=5^2. Another example would be 20^2 and 21^2= 29^2 etc.
Now just to generalise this to n consective integers and powers of cubics:eek: Cheers for the help!

seandoiler

what was the method you used out of interest?

.E_C_K_S.

seandoiler wrote: »

what was the method you used out of interest?

There is a site here which visually shows it for the first few consecutive legs( i.e (3,4),(20,21),(119,120) etc in the The two legs are consecutive section

using a right angled triangle and setting the opposite=1+adjacent (to be consecutive)

Here is the site: http://www.maths.surrey.ac.uk/hosted-sites/R.Knott/Pythag/pythag.html#moremn

For the general form you define the u=x+y and v=y (into formulas on that site) end up with Pells equation and solve for the roots which will give you the nth triple for any n.

seandoiler

WALSHY16291 wrote: »

For the general form you define the u=x+y and v=y (into formulas on that site) end up with Pells equation and solve for the roots which will give you the nth triple for any n.

hmmh not all pells numbers work though (unless i've misunderstood something you've said but i assume that you are suggesting that the pell numbers provide the 'hypothenuese' value for the solution)...

i looked at it a different way, setting [latex]n^2+(n+1)^2=k^2[/latex], solving this yields an equation containingthe term [latex]\sqrt{2k^2-1}[/latex], now for us to have an integer solution, we require that this is an integer, find first couple of examples, [latex]k=1,5,29,169,...[/latex] spot pattern that [latex]k[n]=6*k[n-1]-k[n-2][/latex]...this then generates the longer side, to get the other sides we just note that [latex]n=\lfloor \sqrt{\frac{k^2}{2}} \rfloor[/latex], and of course it's then easy to calculate n+1

the numbers i generate for the k seem to be every second number of the pells numbers that you talk about (from the first couple I've checked at any rate).....what are your opinions on this solution?

.E_C_K_S.

Ya my earlier response may have been confusing because I kind of rushed it:D.

I know that the equation is of the form n^2+(n+1)^2=k^2 but how did you get that equation containing the term? Probably just something that I'm missing out on.
Also when you sub back in that equation for n it does not equal k^2 for me:(

With regards to the pell equation this link should help you understand what I mean. We let the opposite+/-1= adjacent (to be consecutive) and then rearrange to get into a form that is of the pell form

http://mathworld.wolfram.com/TwinPythagoreanTriple.html

seandoiler

[latex] n^2+(n+1)^{2}=k^{2}[/latex]
[latex] 2n^2+2n+1-k^2=0 [/latex]
[latex] n^2+n+\frac{(1-k^2)}{2}=0[/latex]
Solve this then using "-b" formula
[latex]n=\frac{-1 \pm \sqrt{1-4(\frac{1-k^{2}}{2})}}{2}[/latex]
[latex]n=\frac{-1 \pm \sqrt{2k^2-1}}{2}[/latex]
now just need to recognise that [latex]n[/latex] must be an integer, so [latex]\sqrt{2k^2-1}[/latex] must be an integer and then i went from there

.E_C_K_S.

seandoiler wrote: »

[latex] n^2+(n+1)^{2}=k^{2}[/latex]
[latex] 2n^2+2n+1-k^2=0 [/latex]
[latex] n^2+n+\frac{(1-k^2)}{2}=0[/latex]
Solve this then using "-b" formula
[latex]n=\frac{-1 \pm \sqrt{1-4(\frac{1-k^{2}}{2})}}{2}[/latex]
[latex]n=\frac{-1 \pm \sqrt{2k^2-1}}{2}[/latex]
now just need to recognise that [latex]n[/latex] must be an integer, so [latex]\sqrt{2k^2-1}[/latex] must be an integer and then i went from there

Oh I See now:) Ya thats an interesting way to look at it.If I have time I might do both. Did you have a look at that link. I think they both end up with the same general idea anyways!