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Does anyone know of a pair of Pythagorean triples of the form $$(a, b, M) \quad\text{and}\quad(b, b+a, N)$$ Is such a pair possible?

Batominovski
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limepickle
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    $$M^2 = a^2 + b^2$$ $$N^2 = b^2+(b+a)^2 $$ $$N^2 = b^2 + 2ab + a^2 + b^2$$ $$N^2= b^2 + 2ab + M^2$$ $$N > a+b >M > b > a > 0$$ $$***$$ $$a = r^2 - k^2$$ $$b = 2kr$$ $$M = r^2+k^2$$ $$N^2 = 2b^2 + 2ab + a^2$$

    Using these facts, I wrote a Javascript program to check for solutions, and found none for a<10000, b<10000. I'll post the code here for review

     – Brevan Ellefsen Sep 23 '15 at 04:10
  • for (a=1; a<10000; a++) { for (b=1; b<10000; b++) { if (Math.sqrt(2*b*b + 2*a*b + a*a)%1== 0 && Math.sqrt((a*a+b*b))%1==0) { document.write(a + " | " + b + "
    ") } } }     – Brevan Ellefsen Sep 23 '15 at 04:11
  • @BrevanEllefsen would like to know how to get statements displayed on separated lines in a comment. – Mick Sep 23 '15 at 05:01
  • @Mick Same way you do it with regular posts. For example, lets say you have the equation $5x+y$. You can write it as I just did in-line as $5x+y$, or you can write it as $$5x+y$$, yielding $$5x+y$$ – Brevan Ellefsen Sep 23 '15 at 05:03
  • I have ported my code to Java to optimize speed a bit, and I have continued testing. So far my code has returned no solutions when $a<10^5$ and $b<10^5$. I am using effectively the same code as above. If I remove the condition to test for $N$ being a perfect integer I get a list of all Pythagorean triples, and if I remove the Pythagorean triple condition I get a list of all $a$ and $b$ that satisfy the formula for $N^2$, so I'm confident in the quality of my answer (floating point always worries me a tiny bit though...) – Brevan Ellefsen Sep 23 '15 at 05:06
  • These calculations are taking exponentially longer to solve, to the point where my computer has just been chugging away on the problem $a<10^6$, $b<10^6$. Unfortunately I found nothing. That doesn't mean that we have a proof, but at least we know that any solutions are quite large. Perhaps there are some properties of triplets as $a$ and $b$ approach infinity that can use this fact to form a proof? – Brevan Ellefsen Sep 23 '15 at 05:17
  • The existence of positive integers $a$ and $b$ s.t. $(a,b,c)$ and $(b,a+b,d)$ are Pythagorean triples for some positive integers $c$ and $d$ is equivalent to the existence of coprime positive integers $m$ and $n$ of different parities s.t. $$m^4+2m^2n^2+n^4+4m^3n-4mn^3=k^2$$ for some integer $k$. – Batominovski Mar 21 '20 at 15:57

2 Answers2

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Partial solution follows:

First note that if $a$ and $b$ share a common factor, then you can divide both triples by that common factor and the property still holds (e.g. if $a$ and $b$ are both even, then so is $a+b$, and hence the triplets $(a/2, b/2, M/2)$ and $(b/2, \frac{a+b}{2}, N/2)$ should also be Pythaogorean). Hence, without loss of generality, assume that $\gcd(a, b) = 1$, and hence $(a, b, M)$ is a primitive triple. Then $(b, a+b, N)$ is also a primitive triple.

Then assume that the generators of $(a, b, M)$ and $(b, a+b, N)$ are $(x_1, y_1)$ and $(x_2, y_2)$ respectively, i.e. that $a$ and $b$ equal $2x_1y_1$ and $x_1^2-y_1^2$ in some order, and similarly for $b$ and $a+b$. That gives four possible cases to test against, but three of them can very quickly be eliminated:

$a+b$ is always odd (because exactly one of $a$ and $b$ is even), so it must be that $b = 2x_2y_2$ and $a+b = x_2^2 - y_2^2$. Hence $b$ is even, so $a$ is odd, hence $a = x_1^2 - y_1^2$ and $b = 2x_1y_1$, implying that $x_1y_1 = x_2y_2$.

We can then rearrange all of that to eliminate $a$, $b$ and $y_2$, and hence have an expression relating $x_1$, $y_1$ and $x_2$:

$x_2^4 - (x_1^2 - y_1^2 + 2x_1y_1)x_2^2 - x_1^2y_1^2 = 0$

And we note that in this form, it is a quadratic in $x_2^2$, so the triples exist only if that quadratic has at least one solution that is a perfect square. An equivalent equation can be written for $y_2$. So for any given Pythagorean triple, there is a means of testing if it will form a second triple, but I don't know of any way to simplify that or prove that it will never have a solution.

ConMan
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Using Wolfram Alpha to ensure accurate algegra, we start with Euclid's formula (modified) for the OP.

$$A=m^2-n^2\qquad B=2mn+(m^2-n^2)\qquad C=m^2+n^2$$

$$A^2=m^4 - 2 m^2 n^2 + n^4$$

$$B^2=m^4 + 4 m^3 n + 2 m^2 n^2 - 4 m n^3 + n^4$$

$$C^2=m^4 + 2 m^2 n^2 + n^4$$

$$A^2+B^2-C^2=m^4 + 4 m^3 n - 2 m^2 n^2 - 4 m n^3 + n^4=0$$

The only solutions to the last equation are:

$$n =\pm m\qquad n = 2 m \pm \sqrt{5} m\qquad n=m=0$$

$\therefore$ there are no integer solutions except $A,B,C=(0,0,0), (0,2,2), (0,8,8), (0,18,18), ...$

poetasis
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  • Why is $A^2+B^2-C^2=0$ true? – Batominovski Mar 21 '20 at 15:56
  • Subtracting $C^2$ from both sides of $A^2+B^2=C^2$ leaves $0$ on the right side of the equation. – poetasis Mar 21 '20 at 16:01
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    I think you are falsely assuming that if $(m^2-n^2,2mn,m^2+n^2)$ and $(2mn,2mn+m^2-n^2,\text{something})$ are both Pythagorean triples, then $\text{something}$ must equal $m^2+n^2$. – Batominovski Mar 21 '20 at 16:03
  • The sum $A^2+B^2$ does not equal C. It equals $C^2=(m^2+n^2)^2=m^4+2m^2n^2+n^4$ which is what I subtracted from $A^2+B^2$ to get zero. – poetasis Mar 21 '20 at 16:09
  • How is the last comment about $A^2+B^2$ not equaling $C$ relevant to my question? I didn't assume $A^2+B^2=C$. The question is basically this: why is $A^2+B^2=C^2$ in the first place? – Batominovski Mar 21 '20 at 16:10
  • Sorry. I misread. Yes, $something$ is $m^2+n^2$ but the only values of $m&n$ that satisfy the latter are irrational, meaning there is no Pythagorean triple because, by definition, such a triple consists of integers. – poetasis Mar 21 '20 at 16:15
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    I think you are not trying to understand/answer my question. Please explain why $A^2+B^2=C^2$ given $A=m^2-n^2$, $B=2mn+(m^2-n^2)$, and $C=m^2+n^2$. (This is the same as explaining why "$\text{something}$" is $m^2+n^2$.) – Batominovski Mar 21 '20 at 16:18
  • $A^2+B^2\ne C^2$ for a non-trivial triple unless $m$ and/or $n$ are irrational or of opposite sign or zero. For $n=-m$, we have $(0,2,2), (0,8,8), (0,18,18),$ etc. – poetasis Mar 21 '20 at 16:33
  • I think you went so far away from the point of my question, but anyway, somebody may be able to make better arguments than me about why your answer is highly dubious (very likely, it is completely wrong). – Batominovski Mar 21 '20 at 16:35