# Hypergeometric Solution? **URL:** \ **Category:** Factual Questions\ **Created:** [March 6, 2001, 6:58pm UTC](https://boards.straightdope.com/t/hypergeometric-solution/56975 "2001-03-06T18:58:47Z")\ **Posts on this page:** 4\ **Page:** 1
**Author:** ![Enola\_Straight](https://avatars.discourse-cdn.com/v4/letter/e/dec6dc/32.png) [@Enola\_Straight](https://boards.straightdope.com/u/Enola_Straight)\ **Post date:** [March 6, 2001, 6:58pm UTC](https://boards.straightdope.com/t/hypergeometric-solution/56975/1 "2001-03-06T18:58:47Z")
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) ---
**Author:** ![Cabbage](https://avatars.discourse-cdn.com/v4/letter/c/f07891/32.png) [@Cabbage](https://boards.straightdope.com/u/Cabbage)\ **Post date:** [March 6, 2001, 9:05pm UTC](https://boards.straightdope.com/t/hypergeometric-solution/56975/2 "2001-03-06T21:05:24Z")
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](http://br.crashed.net/~akrowne/crc/math/d/d238.htm) to find a list of solutions to your equation. ---
**Author:** ![John\_Kentzel-Griffin](https://avatars.discourse-cdn.com/v4/letter/j/ccd318/32.png) [@John\_Kentzel-Griffin](https://boards.straightdope.com/u/John_Kentzel-Griffin)\ **Post date:** [March 6, 2001, 9:55pm UTC](https://boards.straightdope.com/t/hypergeometric-solution/56975/3 "2001-03-06T21:55:42Z")
[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] ---
**Author:** ![Pies\_R\_Squared](https://avatars.discourse-cdn.com/v4/letter/p/d78d45/32.png) [@Pies\_R\_Squared](https://boards.straightdope.com/u/Pies_R_Squared)\ **Post date:** [March 6, 2001, 10:51pm UTC](https://boards.straightdope.com/t/hypergeometric-solution/56975/4 "2001-03-06T22:51:43Z")
**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