provide integer solutions for a right angled triangle where each side is a perfect square
2006-10-05 00:46:56 · 6 answers · asked by Mein Hoon Na 7 in Science & Mathematics ➔ Mathematics
Can't be done. You see, by the pythagorean theorem, a²+b²=c². Suppose that each of these sides was a perfect square - i.e. a=x², b=y², c=z², for some x, y, z. This would imply that (x²)²+(y²)²=(z²)², that is that x^4+y^4=z^4. However, by Fermat's last theorem, x^4+y^4=z^4 has no integer solutions. Thus, there are no a, b, c fitting your criteria. Q.E.D.
2006-10-05 00:50:38 · answer #1 · answered by Pascal 7 · 3⤊ 0⤋
Can not be done. The most right angles a triangle can have is one. If you add all three angles of a triange they will always add to 180 degrees. If you have one right angel (which is 90 degrees), then you only have another 90 degrees left to divide between the other two angles, so the largest any of the other angles could be is 89 degrees (with 1 degree for the last angle). No way to have three 90 degree angles as that totals 270 degrees.
2006-10-05 01:00:05 · answer #2 · answered by dewcoons 7 · 0⤊ 0⤋
By the Fermat's Great Theorem, there are no integer solutions satisfying x^n + y^n = z^n, where x, and z are all different and n > 2. Since you require the three sides to be perfect squares, then their Pythagorean relationship will become x^4 + y^4 = z^4, which, accdg. to Fermat's Great Theorem, has no integer solutions.
^_^
2006-10-05 01:03:19 · answer #3 · answered by kevin! 5 · 1⤊ 0⤋
properly, assuming you should use any authentic volume length, you will get as on the purpose of a length of 5 as you opt for. 5.a million, 5.01, 5.001... the precision is as a lot as you. If it should be an integer length, then 6 is ideal.
2016-11-26 03:45:04 · answer #4 · answered by Anonymous · 0⤊ 0⤋
Not possible, as indicated above. Sorry.
2006-10-05 00:57:57 · answer #5 · answered by Anonymous · 0⤊ 0⤋
BY DEFINITION....Impossible.
2006-10-05 03:25:06 · answer #6 · answered by bart4play 3 · 0⤊ 0⤋