There is no Pythagorean triple in which the hypotenuse and one leg are the legs of another Pythagorean triple.

Question

According to Wikipedia,

There are no Pythagorean triples in which the hypotenuse and one leg are the legs of another Pythagorean triple.

I cannot find the proof in the citation provided. I am really struggling to proving this, can somebody please direct me to a proof (or give some advice on how to prove it, or prove it themselves). Thanks!

@BrianFitzpatrick, the lengths in the first triangle are switched to the legs in the second. Quite possible over the reals. Start with $(3,4,5),$ next $(4,5, \sqrt {41})$ — Will Jagy, Mar 02 '14 at 19:46
@WillJagy Ahh I mis-interpreted the question. I thought this was stating that "if $c$ is the hypotenuse of a right triangle and $b$ is a leg of that triangle, then there exists a unique right triangle $abc$ with hypotenuse $c$" — Brian Fitzpatrick, Mar 02 '14 at 19:56
The existence of such a pair of triples is equivalent to the statement that both $a^2 + b^2$ and $2a^2 + b^2$ are perfect squares, for some $a$ and $b$. I'm not sure if there is any well known solution to $2a^2 + b^2 = c^2$, though... — Ben Grossmann, Mar 02 '14 at 20:16
@Omnomnomnom, for any prime $p \equiv 1 \pmod 8,$ there are primitive solutions to $x^2 + y^2 = p^2$ and $u^2 + 2 v^2 = p^2.$ However, there will not be much predictable relation among $u,v,x,y.$ — Will Jagy, Mar 02 '14 at 20:20
There are two cases, both very similar to Fermat's proof that there are only trivial solutions to $x^4 + y^4 = z^2,$ meaning either $x$ or $y$ is equal to $0.$ Long but in many books, as it gives the case of exponent 4 in Fermat's Last Theorem. It is generally called the Method of Descent. http://en.wikipedia.org/wiki/Proof_of_Fermat%27s_Last_Theorem_for_specific_exponents#n.C2.A0.3D.C2.A04 and http://en.wikipedia.org/wiki/Proof_by_infinite_descent — Will Jagy, Mar 02 '14 at 22:17
@WillJagy It's about 3/4 of the way down the list of properties on http://en.wikipedia.org/wiki/Pythagorean_triple#Elementary_properties_of_primitive_Pythagorean_triples — Gamma Function, Mar 03 '14 at 00:26
@GammaFunction, thanks, the reference was then Carmichael, R. D., 1914, "Diophantine analysis," in second half of R. D. Carmichael, The Theory of Numbers and Diophantine Analysis, Dover Publ., 1959. — Will Jagy, Mar 03 '14 at 00:33
The proof is available at http://www.gutenberg.org/zipcat.php/20073/20073-pdf.pdf §5 starting on page 10 (pdf page 22) — Neil W, Mar 03 '14 at 01:14
See also: Are there any positive integers $a, b, c, d$ such that both $(a, b, c)$ and $(b, c, d)$ are Pythagorean triples? and Is there $a,b,c,d\in \mathbb N$ so that $a^2+b^2=c^2$, $b^2+c^2=d^2$? — Martin Sleziak, Nov 07 '19 at 20:08
If $A^2=C^2-B^2$, the same B&C and only generate one A-value. — poetasis, Aug 07 '21 at 23:52

score 3 · Answer 1 · edited Apr 13 '17 at 12:21

Oh, well. I cannot find this claim in Wikipedia... one case is easy, the other, not. Any primitive Pythagorean triple is $2xy, x^2 - y^2, x^2 + y^2$ with one of $x,y$ odd and the other even, and $\gcd(x,y) =1.$ We cannot have a Pythagorean triple with legs $x^2 - y^2, x^2 + y^2,$ because we would be solving $$ 2 x^4 + 2 y^4 = z^2 $$ with one of $x,y$ odd. Since that gives us $2 x^4 + 2 y^4 \equiv 2 \pmod 4$ we find immediately that this cannot be an integer square.

the other case is harder, and I do not have it yet. With legs $2xy, x^2 + y^2,$ we are solving $$ \color{magenta}{ x^4 + 6 x^2 y^2 + y^4 = z^2 \; \; \; ?} $$ My little computer run suggests that this is impossible. However....anyway, Fermat's method of descent, which I am not getting to work, is said to be equivalent to something or other using elliptic curves. So i am putting this here and adding that as a tag to the main question.

I think I have the important step, and that there is an elementary book somewhere on elliptic curves with the necessary final step. Mostly from Mordell's book, page 139, we divide the equation in magenta by $y^4,$ replace $x/y$ by $X$ and $z^2/y^4$ by $Z^4,$ after which we are asking about the rational points on $$ \color{magenta}{ X^4 + 6 X^2 + 1 = Z^2 \; \; \; ?} $$ Mordell points out that the points, and especially the rational points, given by the equations $$ S = \frac{X^2 + Z + 1}{2} $$ and then $$ T = 2 X (S+1) $$ take us to the elliptic curve $$ T^2 = 4 S^3 - 4 S, $$ which has no constant term. It is my fervent prayer that this elliptic curve has only the trivial rational points at which $T=0,$ so that the only rational points on the original have $X=0,$ and finally, back to lower case, either $x=0$ or $y=0.$

I WILL FIND OUT

See rational points on particular elliptic curve

score 3 · Answer 2 · edited Apr 13 '17 at 12:19

Suppose that (i) $a^2+b^2=c^2$ and (ii) $b^2+c^2=d^2$. Then, if we divide $(i)$ by $b^2$ we obtain $u^2+1=v^2$ and dividing (ii) by $b^2$ obtain $1+v^2=w^2$, where $u=a/b$, $v=c/b$ and $w=d/b$. In particular $u^2+1=v^2$, $u^2+2=w^2$, and $u^2$ are all squares. Hence, $(u^2,uvw)$ is a point in the elliptic curve $$E:y^2=x(x+1)(x+2).$$ After a change of variables, $x= X-1$, we see that $E$ is isomorphic to $$Y^2=X^3-X=X(X-1)(X+1).$$ This curve only has $4$ rational points, namely $\{\infty,(0,0),(1,0),(-1,0)\}$, so the only points on $E$ are $\{\infty,(0,0),(-1,0),(-2,0)\}$, and therefore the only solutions of (i)+(ii) are trivial.

There is no Pythagorean triple in which the hypotenuse and one leg are the legs of another Pythagorean triple.

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