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Consider a right-angled triangle with integer-valued sides $a<b<c$ where a,b,c are pairwise co-prime. Let $d=c-b$. Suppose $d$ divides $a$. Then

a. Prove that $d \leq 2$

b. Find all such triangles (i.e:) all possible triplets a,b,c) with perimeter less than 100

Source

My attempt:

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

Then,

$$ a^2 = (c-b)(c+b) \tag{1}$$

From the question $a=(c-b)p$ where $p$ is integer:

$$ p^2 (c-b) = (c+b)$$

Now, $c-b$ is a factor of $c+b$

Another observation that I made: It must be that not all sides are odd, this can be concluded from (1), further 'a' can't be odd with $b,c$ even and a,b can't be both even (this would violate the pairwise co-prime criteria as c is also even)

Now... I'm stuck..

tryst with freedom
  • 11,538

1 Answers1

1

Hints for part a:

Show that

  1. $ d \mid c+b$. (Already done by OP)

2. $ d \mid 2c$.

3. $ d \mid 2a$.

3. Since $ \gcd (c,a) = 1$, and using the above, hence $d = 1, 2 $.

Hints for part b:

  1. Recall that the primitive pythagorean triple solution is $ ( m^2 - n^2, 2mn, m^2 + n^2) $.

2. If $ c-b = 1$, show that $(a, b, c) = 2n+1 , 2n^2 + 2n , 2n^2 + 2n + 1 $.

3. If $ c - b = 2$, show that $(a, b, c ) = (2m, m^2 - 1, m^2 + 1)$.

Calvin Lin
  • 68,864
  • I am not sure if I am making mistake but from the way I found 2c can be divided by 'd' , I also found d divides $2b$ , is this right or am I losing my mind? Btw do you have any suggestions where I should study the parameterization of the triplets? I couldn't find no good resources easily – tryst with freedom Jul 05 '21 at 08:50
  • @Buraian 1) You are correct. $ d \mid c+b, d \mid c-b \Rightarrow d\mid (c+b)\pm(c-b) = 2c, 2b$. $\quad$ 2) There can be many ways to prove a fact. You don't necessarily need to follow the path that someone lays out. $\quad$ 3) Wikipedia gives a proof of it - https://en.wikipedia.org/wiki/Pythagorean_triple. It's such a well known fact, I believe you can find other resources for it too. – Calvin Lin Jul 05 '21 at 10:11