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It turns out that there are none zero integer (whole numbers with no decimal points) solutions satisfying 2²=²+², where z is the hypothenuse and x and y are the other sides. For example, z= 5, x = 3 and y=4 works because 5² = 32+ 42. Actually, there are so many other integer solutions to that equation. Try z=10, x=6 and y = 8 1) Give two more integer solutions of the equation = x+y². Remember that you are only allowed to work with non-zero integers (whole numbers- no fractions, no decimal point numbers) = 2) Now, is it possible to extend the Pythagorean Theorem to the third power to become z³ = x+y where z,x and y are whole numbers? In other words, can you find three non-zero integers (z, x and y) satisfying z³ x+y? If so, then you are done from this project. Just write down the integer values of your z, x and y. (To be honest with you, I tried, and I couldn't find any solution). But maybe you are lucky here!

It turns out that there are none zero integer (whole numbers with no decimal points) solutions satisfying 2²=²+², where z is the hypothenuse and x and y are the other sides. For example, z= 5, x = 3 and y=4 works because 5² = 32+ 42. Actually, there are so many other integer solutions to that equation. Try z=10, x=6 and y = 8 1) Give two more integer solutions of the equation = x+y². Remember that you are only allowed to work with non-zero integers (whole numbers- no fractions, no decimal point numbers) = 2) Now, is it possible to extend the Pythagorean Theorem to the third power to become z³ = x+y where z,x and y are whole numbers? In other words, can you find three non-zero integers (z, x and y) satisfying z³ x+y? If so, then you are done from this project. Just write down the integer values of your z, x and y. (To be honest with you, I tried, and I couldn't find any solution). But maybe you are lucky here!

Intermediate Algebra
Intermediate Algebra
10th Edition
ISBN:9781285195728
Author:Jerome E. Kaufmann, Karen L. Schwitters
Publisher:Jerome E. Kaufmann, Karen L. Schwitters
Chapter6: Quadratic Equations And Inequalities
Section6.4: Quadratric Formula
Problem 61PS
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Give two more integer solutions of the equation z2 = x2 + y2 with non-zero integers
is the sum of the squares of the other sides It turns out that there are none zero integer (whole numbers with no decimal points) solutions satisfying 2=²+², where z is the hypothenuse and x and y are the other sides. For example, z= 5, x = 3 and y=4 works because 52 = 3² + 42. Actually, there are so many other integer solutions to that equation Try z=10, x=6 and y = 8 1) Give two more integer solutions of the equation z² = x+y². Remember that you are only allowed to work with non-zero integers (whole numbers- no fractions, no decimal point numbers) = 2) Now, is it possible to extend the Pythagorean Theorem to the third power to become z³ = x³+y³ where z,x and y are whole numbers? In other words, can you find three non-zero integers (z, x and y) satisfying z³ x+y³? If so, then you are done from this project. Just write down the integer values of your z, x and y. (To be honest with you, I tried, and I couldn't find any solution). But maybe you are lucky here! 3) If you couldn't find a solution
Transcribed Image Text:is the sum of the squares of the other sides It turns out that there are none zero integer (whole numbers with no decimal points) solutions satisfying 2=²+², where z is the hypothenuse and x and y are the other sides. For example, z= 5, x = 3 and y=4 works because 52 = 3² + 42. Actually, there are so many other integer solutions to that equation Try z=10, x=6 and y = 8 1) Give two more integer solutions of the equation z² = x+y². Remember that you are only allowed to work with non-zero integers (whole numbers- no fractions, no decimal point numbers) = 2) Now, is it possible to extend the Pythagorean Theorem to the third power to become z³ = x³+y³ where z,x and y are whole numbers? In other words, can you find three non-zero integers (z, x and y) satisfying z³ x+y³? If so, then you are done from this project. Just write down the integer values of your z, x and y. (To be honest with you, I tried, and I couldn't find any solution). But maybe you are lucky here! 3) If you couldn't find a solution
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