This relationship leads to the conclusion that there is no Pythagorean triple of this form that contains the integer 2. Why?

suppose 2^2 + b^2 = c^2 with 2 < b < c (it should be clear that this assumption is fine, 12 + 22 = 5 and sqrt(5) is not an integer; 22 + 22 = 8 and sqrt(8) is not an integer). Then c^2 - 4 = b^2 so (c+2)(c-2) = b^2. As b is prime, this means c-2=b or c-2=1 (since c+2 > 1). So either c=b+2 or c=3. The c=3 case can't happen since that would make b^2 = 5 so we are left with c=b+2. Then we have 4 + b^2 = (b+2)2 = b2 + 4b + 4 meaning that b=0. So there are no triples involving 2.

Step-by-step explanation:

onyebuchinnaji onyebuchinnaji · Answer 2 · 2021-09-16T15:10:19+02:00

A Pythagorean triple of this form cannot contain integer 2 if we add extra 2x²y² because it will increase the value of the coefficient of xy to 4.

The given function:

[tex](x^2-y^2) + (2xy)^2 = (x^2 + y^2)^2[/tex]

This function is expanded as follows;

[tex](x^2-y^2)^2 + (2xy)^2 = (x^2 + y^2)^2\\\\x^4 + y^4 -2x^2y^2 + 4x^2y^2 = x^4 +y^4 + 2x^2y^2\\\\add \ \ (+2x^2y^2) \ to \ both \ sides \\\\x^4 +y^4 +4x^2y^2 = x^4 +y^4 +4x^2y^2[/tex]

In this expanded form of the given function, it is clear that non of the coefficients of x and y contains integer 2 and the supposed coefficient value of xy increase to 4 because of the extra 2x²y² added to the function.

Thus, we can conclude that a Pythagorean triple of this form cannot contain integer 2 if we add extra 2x²y² because it will increase the value of the coefficient of xy to 4.

Learn more here: https://brainly.com/question/11279692