On some Fermat triangles

Question

Fermat noticed that in the right triangle $(156, 1517, 1525)$ the square of the difference of the legs exceeds the double of the square of the least leg by a square, so being $(a, b, c)$, other than $a^2 + b^2 = c^2$, the three sides of this right triangle also obey,

$$(a - b)^2 - 2b^2 = d^2$$

For those kind of triangles, one can prove that

$$2c^4 - d^4 = g^2\tag1$$

where $g = (a+b)^2 - 2b^2$.

To compare, the quartic Diophantine equation,

$$2z^4-y^4 = x^2\tag2$$

has infinitely many solutions:

$(x, y, z) = (1, 1, 1), (239, 1, 13), (2750257, \color{red}{1343}, \color{blue}{1525}), (3503833734241, 2372159, 2165017), (2543305831910011724639, 9788425919, 42422452969), (76285433470805578504147559981041, \color{red}{5705771236038721}, \color{blue}{7658246457672229}),(\small{21823524993202203117598843430945898516918043105969513349188339015638198841921},\color{red}{75727152767742719949099952561135816319},\color{blue}{126314830357375266295717376544111167953})$

So now is possible to find the sides of those infinite Fermat triangles:

$(a, b, c, d) = (1517, 156, \color{blue}{1525}, \color{red}{1343}), (7518954988589021, 1453978915260300, \color{blue}{7658246457672229}, \color{red}{5705771236038721}), (121931332515412369802346183394249940305, 32987672231129530249809032005100126928, \color{blue}{126314830357375266295717376544111167953}, \color{red}{75727152767742719949099952561135816319})$

I noticed that some third solution (the $z$) of the quartic represent a solution (the $c$) of those Fermat triangles. Why is that?

Answer 1

If you compare the relationship obeyed by that special Pythagorean triple,

$$2c^4 - d^4 = g^2\tag1$$

with the quartic Diophantine equation you cited (and which I rearranged for clarity),

$$2z^4-y^4 = x^2\tag2$$

then it now becomes obvious that $c = z$, and $d = y$, so they were just essentially the same relation.

However, since $a^2+b^2 = c^2$ allows for negative solutions $(\pm a, \pm b)$, then that is why you were missing some Pythagorean triples. For example, your second quartic solution above,

$$(x, y, z) = (3503833734241, \color{red}{2372159}, \color{blue}{2165017})$$

leads to the Pythagorean triple (plus $d$),

$$(a,b,c,d) = (2150905, -246792, \color{blue}{2165017}, \color{red}{2372159})$$

which is missing from your list.

Answer 2

This is about integer solutions to the system of Diophantine equations

$$ a^2 + b^2 = c^2, \tag1 $$ $$ (a-b)^2 - 2 b^2 = d^2, \tag2 $$ $$ 2 c^4 - d^4 = g^2, \tag3 $$ $$ 2 z^4 - y^4 = x^2. \tag4 $$

First, verify the algebraic identity

$$ e^2 + g^2 = 2(a^2+b^2)^2 \; \text{ where } \\ e = (a-b)^2 - 2b^2,\;\; g = (a+b)^2 - 2b^2. \tag5 $$

Compare this to equations $(1)$,$(2)$,$(3)$ implying that $\,e = d^2\,$ for some integer $\,d.\,$

Euclid's formula for Pythagorean triples applied to equation $(1)$ is

$$ a = u^2 - v^2, \qquad b = 2uv, \qquad c = u^2 + v^2. \tag6 $$

Substitute this in equation $(2)$ to get a quartic Diophantine equation

$$ e = d^2 = u^4-4 u^3 v-6 u^2 v^2+4 u v^3+v^4. \tag7 $$

One method of solving these is to use some generalized Somos-4 sequences.

More precisely, define the integer sequences

$$ e_0 = 0,\, e_1 = 1,\, e_2 = 2,\, e_3 = -13,\, e_4 = 84,\, e_5 = 1525,\\ e_n = (-13\,e_{n-1}e_{n-4} -42\,e_{n-2}e_{n-3})/e_{n-5}. $$ $$ f_0 = 1,\, f_1 = 1,\, f_2 = -3,\, f_3 = 1,\, f_4 = 113,\\ f_n = (-13\,f_{n-1}f_{n-4} -42\,f_{n-2}f_{n-3})/f_{n-5}. $$ $$ g_0 = 1,\, g_1 = -1,\, g_2 = -7,\, g_3 = 239,\\ g_n = (36\,g_{n-1}g_{n-3} +13\,g_{n-2}g_{n-2})/g_{n-4}, $$ $$ u_n = e_n f_{n-1},\quad v_n = f_n e_{n-1}, $$ $$ b_n = 2(-1)^n u_n v_n,\quad a_n = u_n^2 - v_n^2. $$

Their first few values are

$$ e = (0, 1, 2, -13, 84, 1525, 6214, -2165017, -151245528, 42422452969,\dots), $$ $$ f = (1, 1, -3, 1, 113, -1343, -57123, -2372159, 262621633, -9788425919,\dots), $$ $$ g = (1, -1, -7, 239, -7967, -2750257, 3262580153, -3503833734241,\dots), $$ $$ u = (0, -3, 2, -1469, -112812, -87112575, -14740596026, -568580300012761,\dots), $$ $$ v = (1, 2, 39, 84, 172325, -8345402, 123672266091, 358778440454952,\dots), $$ $$ b = (0, -12, -156, -246792, 38880655800, 1453978915260300,\dots), $$ $$ a = (-1, 5, -1517, 2150905, -16969358281, 7518954988589021,\dots). $$

Your table of values of $\,(x, y, z)\,$ are the values of $\,(g_{2n+1}, f_{2n+1}, e_{2n+1}).$

Your table of values of $\,(a, b, c, d)\,$ are the values of $\,(a_{3n}, b_{3n}, e_{6n-1}, f_{6n-1}).$

This answers your question:

I noticed that some third solution (the z) of the quartic represent a solution (the c) of those Fermat triangles. Why is that?

Now you can verify that equations $(1)$ through $(4)$ are satisfied except in equation $(3)$ replace $\,g\,$ with $\,x\,$.

Compare this to my answer to MSE question 2972590 about the same Fermat Diophantine equation.