Solve 1/2(a+b)^2+1/2(a-b)^2=a^2+b^2

$$\frac { 1 } { 2 } ( a + b ) ^ { 2 } + \frac { 1 } { 2 } ( a - b ) ^ { 2 } = a ^ { 2 } + b ^ { 2 }$$

Solve for a (complex solution)

$$a\in \mathrm{C}$$

Solution Steps

Use binomial theorem \left(p+q\right)^{2}=p^{2}+2pq+q^{2} to expand \left(a+b\right)^{2}.

$$\frac{1}{2}\left(a^{2}+2ab+b^{2}\right)+\frac{1}{2}\left(a-b\right)^{2}=a^{2}+b^{2}$$

Use the distributive property to multiply \frac{1}{2} by a^{2}+2ab+b^{2}.

$$\frac{1}{2}a^{2}+ab+\frac{1}{2}b^{2}+\frac{1}{2}\left(a-b\right)^{2}=a^{2}+b^{2}$$

Use binomial theorem \left(p-q\right)^{2}=p^{2}-2pq+q^{2} to expand \left(a-b\right)^{2}.

$$\frac{1}{2}a^{2}+ab+\frac{1}{2}b^{2}+\frac{1}{2}\left(a^{2}-2ab+b^{2}\right)=a^{2}+b^{2}$$

Use the distributive property to multiply \frac{1}{2} by a^{2}-2ab+b^{2}.

$$\frac{1}{2}a^{2}+ab+\frac{1}{2}b^{2}+\frac{1}{2}a^{2}-ab+\frac{1}{2}b^{2}=a^{2}+b^{2}$$

Combine \frac{1}{2}a^{2} and \frac{1}{2}a^{2} to get a^{2}.

$$a^{2}+ab+\frac{1}{2}b^{2}-ab+\frac{1}{2}b^{2}=a^{2}+b^{2}$$

Combine ab and -ab to get 0.

$$a^{2}+\frac{1}{2}b^{2}+\frac{1}{2}b^{2}=a^{2}+b^{2}$$

Combine \frac{1}{2}b^{2} and \frac{1}{2}b^{2} to get b^{2}.

$$a^{2}+b^{2}=a^{2}+b^{2}$$

Subtract a^{2} from both sides.

$$a^{2}+b^{2}-a^{2}=b^{2}$$

Combine a^{2} and -a^{2} to get 0.

$$b^{2}=b^{2}$$

Reorder the terms.

$$\text{true}$$

This is true for any a.

$$a\in \mathrm{C}$$

Solve for b (complex solution)

$$b\in \mathrm{C}$$

Solution Steps

Use binomial theorem \left(p+q\right)^{2}=p^{2}+2pq+q^{2} to expand \left(a+b\right)^{2}.

$$\frac{1}{2}\left(a^{2}+2ab+b^{2}\right)+\frac{1}{2}\left(a-b\right)^{2}=a^{2}+b^{2}$$

Use the distributive property to multiply \frac{1}{2} by a^{2}+2ab+b^{2}.

$$\frac{1}{2}a^{2}+ab+\frac{1}{2}b^{2}+\frac{1}{2}\left(a-b\right)^{2}=a^{2}+b^{2}$$

Use binomial theorem \left(p-q\right)^{2}=p^{2}-2pq+q^{2} to expand \left(a-b\right)^{2}.

$$\frac{1}{2}a^{2}+ab+\frac{1}{2}b^{2}+\frac{1}{2}\left(a^{2}-2ab+b^{2}\right)=a^{2}+b^{2}$$

Use the distributive property to multiply \frac{1}{2} by a^{2}-2ab+b^{2}.

$$\frac{1}{2}a^{2}+ab+\frac{1}{2}b^{2}+\frac{1}{2}a^{2}-ab+\frac{1}{2}b^{2}=a^{2}+b^{2}$$

Combine \frac{1}{2}a^{2} and \frac{1}{2}a^{2} to get a^{2}.

$$a^{2}+ab+\frac{1}{2}b^{2}-ab+\frac{1}{2}b^{2}=a^{2}+b^{2}$$

Combine ab and -ab to get 0.

$$a^{2}+\frac{1}{2}b^{2}+\frac{1}{2}b^{2}=a^{2}+b^{2}$$

Combine \frac{1}{2}b^{2} and \frac{1}{2}b^{2} to get b^{2}.

$$a^{2}+b^{2}=a^{2}+b^{2}$$

Subtract b^{2} from both sides.

$$a^{2}+b^{2}-b^{2}=a^{2}$$

Combine b^{2} and -b^{2} to get 0.

$$a^{2}=a^{2}$$

Reorder the terms.

$$\text{true}$$

This is true for any b.

$$b\in \mathrm{C}$$

Solve for a

$$a\in \mathrm{R}$$

Solution Steps

Use binomial theorem \left(p+q\right)^{2}=p^{2}+2pq+q^{2} to expand \left(a+b\right)^{2}.

$$\frac{1}{2}\left(a^{2}+2ab+b^{2}\right)+\frac{1}{2}\left(a-b\right)^{2}=a^{2}+b^{2}$$

Use the distributive property to multiply \frac{1}{2} by a^{2}+2ab+b^{2}.

$$\frac{1}{2}a^{2}+ab+\frac{1}{2}b^{2}+\frac{1}{2}\left(a-b\right)^{2}=a^{2}+b^{2}$$

Use binomial theorem \left(p-q\right)^{2}=p^{2}-2pq+q^{2} to expand \left(a-b\right)^{2}.

$$\frac{1}{2}a^{2}+ab+\frac{1}{2}b^{2}+\frac{1}{2}\left(a^{2}-2ab+b^{2}\right)=a^{2}+b^{2}$$

Use the distributive property to multiply \frac{1}{2} by a^{2}-2ab+b^{2}.

$$\frac{1}{2}a^{2}+ab+\frac{1}{2}b^{2}+\frac{1}{2}a^{2}-ab+\frac{1}{2}b^{2}=a^{2}+b^{2}$$

Combine \frac{1}{2}a^{2} and \frac{1}{2}a^{2} to get a^{2}.

$$a^{2}+ab+\frac{1}{2}b^{2}-ab+\frac{1}{2}b^{2}=a^{2}+b^{2}$$

Combine ab and -ab to get 0.

$$a^{2}+\frac{1}{2}b^{2}+\frac{1}{2}b^{2}=a^{2}+b^{2}$$

Combine \frac{1}{2}b^{2} and \frac{1}{2}b^{2} to get b^{2}.

$$a^{2}+b^{2}=a^{2}+b^{2}$$

Subtract a^{2} from both sides.

$$a^{2}+b^{2}-a^{2}=b^{2}$$

Combine a^{2} and -a^{2} to get 0.

$$b^{2}=b^{2}$$

Reorder the terms.

$$\text{true}$$

This is true for any a.

$$a\in \mathrm{R}$$

Solve for b

$$b\in \mathrm{R}$$

Solution Steps

Use binomial theorem \left(p+q\right)^{2}=p^{2}+2pq+q^{2} to expand \left(a+b\right)^{2}.

$$\frac{1}{2}\left(a^{2}+2ab+b^{2}\right)+\frac{1}{2}\left(a-b\right)^{2}=a^{2}+b^{2}$$

Use the distributive property to multiply \frac{1}{2} by a^{2}+2ab+b^{2}.

$$\frac{1}{2}a^{2}+ab+\frac{1}{2}b^{2}+\frac{1}{2}\left(a-b\right)^{2}=a^{2}+b^{2}$$

Use binomial theorem \left(p-q\right)^{2}=p^{2}-2pq+q^{2} to expand \left(a-b\right)^{2}.

$$\frac{1}{2}a^{2}+ab+\frac{1}{2}b^{2}+\frac{1}{2}\left(a^{2}-2ab+b^{2}\right)=a^{2}+b^{2}$$

Use the distributive property to multiply \frac{1}{2} by a^{2}-2ab+b^{2}.

$$\frac{1}{2}a^{2}+ab+\frac{1}{2}b^{2}+\frac{1}{2}a^{2}-ab+\frac{1}{2}b^{2}=a^{2}+b^{2}$$

Combine \frac{1}{2}a^{2} and \frac{1}{2}a^{2} to get a^{2}.

$$a^{2}+ab+\frac{1}{2}b^{2}-ab+\frac{1}{2}b^{2}=a^{2}+b^{2}$$

Combine ab and -ab to get 0.

$$a^{2}+\frac{1}{2}b^{2}+\frac{1}{2}b^{2}=a^{2}+b^{2}$$

Combine \frac{1}{2}b^{2} and \frac{1}{2}b^{2} to get b^{2}.

$$a^{2}+b^{2}=a^{2}+b^{2}$$

Subtract b^{2} from both sides.

$$a^{2}+b^{2}-b^{2}=a^{2}$$

Combine b^{2} and -b^{2} to get 0.

$$a^{2}=a^{2}$$

Reorder the terms.

$$\text{true}$$

This is true for any b.

$$b\in \mathrm{R}$$