# Hypergeometric Solution?

Everyone’s heard of the Pythagorean theorem:
X^2 + Y^2 + Z^2 (ex. 3^2 + 4^2 = 5^2…9+16=25)
The solution is a whole number…not a fraction or decimal;
this is an important point in my question.

Once, while doodling around, I discovered:
3^3 + 4^3 + 5^3 = 6^3…27+64+125=216

Unfortunately, taken to the next level:
3^4 + 4^4 + 5^4 + 6^4 DOES NOT = 7^4

Does anyone know of a solution to:
A^4 + B^4 + C^4 + D^4 = E^4,
where A,B,C,D,and E are whole numbers?

(Drink ice water to prevent encephalic combustion)

Back in the 19th century, Euler conjectured some stuff about these types of equations. First, you’ve got the Pythagorean Theorem, a[sup]2[/sup]+b[sup]2[/sup]=c[sup]2[/sup].

Next, you’ve got Fermat’s Last Theorem that says you can’t have a[sup]3[/sup]+b[sup]3[/sup]=c[sup]3[/sup]. You can, however (as you noticed), have a[sup]3[/sup]+b[sup]3[/sup]+c[sup]3[/sup]=d[sup]3[/sup].

Euler thought you would have to add at least n nth powers to have the result be an nth power.

He was wrong, though. In 1987 or so, this was found:

2682440[sup]4[/sup]+15365639[sup]4[/sup]+18796760[sup]4[/sup]=20615673[sup]4[/sup].

As for your question, you can check here to find a list of solutions to your equation.

[nitpick]
Pythagorean theorem says that for a right triangle with two sides of a and b with a hypotenuse of c we have:
a[sup]2[/sup] + b[sup]2[/sup] = c[sup]2[/sup]. Where a, b, and c are real numbers.

There happen to be solutions to this among the integers, but the Pythagorean theorem just requires that a, b, and c be real.

An equation whose solutions come from the integers is a Diophantine equation.
[/nitpick]

enolancooper,

From your examples, it seems also that you’re wanting to know if there is an integer x such that x^4 + (x+1)^4 +(x+2)^4 +(x+3)^4 = (x+4)^4.

As you stated, x^2 +(x+1)^2 = (x+2)^2 has a solution of x=3. There is one other solution, however; namely, x=-1.

Next, looking at the equation x^3 + (x+1)^3 + (x+2)^3 = (x+3)^3, after some simplification, we have the cubic x^3=6x + 9. This can be solved by making the substitution x=a+b and so on, but since we already know 3 is a root, we can factor x-3 from x^3 - 6x - 9. This results in (x-3)(x^2 + 3x +3), so clearly 3 is the only real root to the equation being considered.

Now, for x^4 + … + (x+3)^4 = (x+4)^4. This simplifies to 3x^4 + 8x^3 - 12x^2 - 112x - 158 = 0 (if I didn’t make any errors). As a consequence of the ‘rational root test’, the only possible integral solutions of this equation are +/-1, +/-2, +/-79, and +/-158. None of these are roots.

Sincerely,
Pies ‘R’ Squared