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I. The Jacobi-Madden equation, $$a^4+b^4+c^4+d^4 = (a+b+c+d)^4$$

is equivalent to a disguised Pythagorean triple, $$(a^2+ab+b^2)^2+(c^2+cd+d^2)^2 = \big((a+b)^2+(a+b)(c+d)+(c+d)^2\big)^2$$


II. A special case of the Descartes' circle theorem,

$$2(a^4+b^4+c^4+d^4)=(a^2+b^2+c^2+d^2)^2$$

and Euler showed this is just,

$$(2ab)^2+(2cd)^2 = (a^2+b^2-c^2-d^2)^2$$


III. The Fermat quartic,

$$a^4+b^4 = c^4$$

becomes,

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


IV. The Pell equation,

$$x^2-2y^2 = -1$$

is also,

$$x^2 + (y^2 - 1)^2 = y^4$$

as well as,

$$\Big(\frac{x-1}{2}\Big)^2+\Big(\frac{x+1}{2}\Big)^2 = y^2$$

with the latter showing there are infinitely many triples where the legs differ by just $1$.

Q: Are there any other examples of simple quadratic or quartic equations that can be expressed as a Pythagorean triple?

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  • $\begingroup$ The key problem is to explain precisely what "transformed" and "equivalent" mean. Then move on to "simple". $\endgroup$ – Somos Jan 3 '18 at 19:39
  • $\begingroup$ @Somos: Given a pair of equations, if they are expressed as $F(x)=0$ and the $\text{LHS}$ expanded, then the pair are equivalent if the expansion results in exactly the same expression. I've also changed the word "transformed" to "expressed". $\endgroup$ – Tito Piezas III Jan 4 '18 at 2:16
  • $\begingroup$ What if we start with some second-degree (or larger degrees) polynomials in several variables and then square them to see when we can obtain fourth degree ones (or larger ones) that are squares, that is, by going in reverse, would that accomplish something? This is then a polynomial theory problem. $\endgroup$ – user480281 Jan 4 '18 at 2:20
  • $\begingroup$ @Adeen: That is possible, as long as the resulting fourth degree expression doesn't look like mess. $\endgroup$ – Tito Piezas III Jan 4 '18 at 2:26
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    $\begingroup$ Hahah it could look like a mess, but these expressions in your question are short, I believe that you would like to see some huge ones. $\endgroup$ – user480281 Jan 4 '18 at 2:31
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Not sure if this is what you had in mind, but suppose we have two distinct triangular numbers whose product is a square. Say $[u(u+1)/2][v(v+1)/2] = t^2$, where $u > v > 0$. Define $a = 4t$, $b_1 = u - v$, $b_2 = u + v + 1$, $c = 2uv + u + v$. Then a routine calculation gives $$a^2 + b_1^2 = c^2,\quad a^2 + b_2^2 = (c + 1)^2.$$ Thus we get two Pythagorean triangles with a common side and hypotenuses differing by $1$. For instance $u = 8$, $v = 1$ gives $(24, 7, 25)$ and $(24, 10, 26)$. Alternatively you can start with two such triangles and reverse the procedure, as is easily proved.

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  • $\begingroup$ Not quite what I had in mind. But interesting, nonetheless. $\endgroup$ – Tito Piezas III Jan 4 '18 at 14:27

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