Use the polynomial identity (x^(2)+y^(2))2=(x^(2)-y^(2))2+(2xy)2 to generate a Pythagorean triple when x equals 7 and y equals 3. Which of the following is one of the values of the Pythagorean triple?

Answer 1

To use the polynomial identity $(x^2+y^2{)}^2=(x^2-y^2{)}^2+(2xy{)}^2$ to generate a Pythagorean triple, let's start by substituting $x=7$ and $y=3$.

First, calculate $x^2$ and $y^2$: $$x^2=7^2=49$$ $$y^2=3^2=9$$

Now, we can calculate $x^2+y^2$: $$x^2+y^2=49+9=58$$

Next, we calculate $(x^2+y^2{)}^2$: $$(x^2+y^2{)}^2={58}^2=3364$$

Now, calculate $x^2-y^2$: $$x^2-y^2=49-9=40$$

Next, calculate $(x^2-y^2{)}^2$: $$(x^2-y^2{)}^2={40}^2=1600$$

Now, calculate $2xy$: $$2xy=2\cdot 7\cdot 3=42$$

Then calculate $(2xy{)}^2$: $$(2xy{)}^2={42}^2=1764$$

To verify the identity, check that $(x^2-y^2{)}^2+(2xy{)}^2$ equals $(x^2+y^2{)}^2$: $$(x^2-y^2{)}^2+(2xy{)}^2=1600+1764=3364$$

This confirms the identity holds true.

Now we summarize the Pythagorean triple generated by the calculations:

• $a=x^2-y^2=40$
• $b=2xy=42$
• $c=x^2+y^2=58$

Thus, the generated Pythagorean triple is $(40,42,58)$.

Among these values, one of the values of the Pythagorean triple is 40.