# Pythagorean Triplets with “Bounds”

I am interested in the algebraic/geometric way of finding the pythagorean triplets such that

$$a^2 + b^2 = c^2$$

$$a + b + c = 1000$$

I do the obvious

$$a + b = 1000 - (a^2 + b^2)^{1/2}$$

$$a^2 + b^2 = 1000^2 -2(1000)a - 2(1000)b +2ab + a^2 +b^2$$

$$2a + 2b - \frac{2ab}{1000} = 1000$$

$$a + b -\frac{ab}{1000} = 500$$

I have no idea what to do at this point. I can't separate the variables, and any geometric solution is beyond my reach. The only other thing I can think of is writing

$$a + b = 500 + \frac{ab}{1000} = 1000 - c$$

But this is going backwards

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Only 1 exists right? look at answers.yahoo.com/question/index?qid=20091110033358AAfNaHy and Euclids formula en.wikipedia.org/wiki/Pythagorean_triple – mathguy Aug 7 '12 at 16:38
Do you want to include solutions with negative values like $a=1500$, $b=2000$, and $c=-2500$? – Byron Schmuland Aug 7 '12 at 17:01

The following is copy and pasted directly from Yahoo Answers

All Pythagorean triples are generated by ${m^2+n^2, m^2-n^2, 2mn}$, where $m$ and $n$ are positive integers, and $m\gt n$.

You need $a+b+c=1000$, yielding $m(m+n)=500$. So, $m$ and $(m+n)$ are factors of $500$.

$m+n\gt m$, so $m(m+n)\gt m^2$, so $m\lt \sqrt{(500)}$, and $m+n\lt2m$ (since $m\gt n$), so $m(m+n)\lt2m^2$, so $m\gt \sqrt{(250)}$.

The prime factorisation of 500 is $2 \cdot 2 \cdot 5 \cdot 5 \cdot 5$, so the only factor of $500$ between $15 \cdot 8$ and $22 \cdot 4$ is $20$. Thus, $m=20$, $n=5$, giving the answer required: $m^2+n^2$ $= 425$; $m^2-n^2 = 375$; $2mn = 200$.

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We know the parametric form of the Pythagorian triplet is $(k(p^2-q^2), 2kpq, k(p^2+q^2))$ where p,r integers.

So, here $k(p^2-q^2+ 2pq + p^2+q^2)=1000$

=>$kp(p+q)=500$

Clearly, k can be any divisor of 500.

If k=1,

If p=1, p+q=500=> q = 499,

if p=2, p+q=250=> q = 248 and so on.

If k=2,$p(p+q)=250$

If k=5,$p(p+q)=100$

If k=10, $p(p+q)=50$ and so on.

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You don't get all of the non-primitive triples this way (e.g. $9^2 + 12^2 = 15^2$). – Qiaochu Yuan Aug 7 '12 at 16:43
@QiaochuYuan, I've rectified the answer and I think the result will be exhaustive if negative values are permitted. – lab bhattacharjee Aug 7 '12 at 18:00