Q. \(a^2 + b^2 = c\).

Q: What does a^2 + b^2 = c represent geometrically for real a,b,c?

For real numbers with c>0, it's the circle centered at the origin with radius \sqrt{c}. If c=0 it's the origin point; if c<0 there are no real solutions.

Q: When do integers a,b exist for a given integer c?

An integer c is a sum of two integer squares iff in its prime factorization every prime p\equiv 3\pmod 4 appears with an even exponent (classical sum-of-two-squares theorem).

Q: Can a prime p be written as a^2+b^2?

Yes exactly when p=2 or p\equiv 1\pmod 4. Primes p\equiv 3\pmod 4 cannot be expressed as a sum of two integer squares.

Q: How do you solve for b given a and c?

Solve b=\pm\sqrt{\,c-a^2\,}. For real b require c-a^2\ge 0; for integer b require c-a^2 to be a perfect square.

Q: Is there a simple parametrization of all integer solutions (a,b,c)?

There is no single simple parametrization for a^2+b^2=c with arbitrary integer c. For fixed c you search integer pairs; for the related equation a^2+b^2=c^2 there is a classical Pythagorean parametrization.

Q: How many integer representations can a fixed c have?

How many integer representations can a fixed c have?

Q: What parity or modular restrictions exist for a^2+b^2?

Squares are 0 or 1\pmod 4, so a^2+b^2 can be 0,1, or 2\pmod 4. In particular it cannot be 3\pmod 4. If both a,b are odd then a^2+b^2\equiv 2\pmod 4.

Q: How do Gaussian integers help with a^2+b^2=c?

Write c=(a+bi)(a-bi) in the Gaussian integers. Unique factorization there translates the sum-of-two-squares problem into factoring primes and exponents, explaining existence and counting of representations.

Answer

From \(a^2 + b^2 = c\):

Explanation: isolate the desired variable and take square roots (with sign and domain conditions).

Final results:
\(c = a^2 + b^2\)

\(a = \pm\sqrt{c – b^2},\) valid when \(c – b^2 \ge 0\)

\(b = \pm\sqrt{c – a^2},\) valid when \(c – a^2 \ge 0\)

Detailed Explanation

Problem

Solve the equation given by the relation

\[ a^2 + b^2 = c \]

Step-by-step solution and detailed explanations

Isolate the variable \( c \).

The relation is already solved for \( c \) in terms of \( a \) and \( b \). Writing it explicitly:

\[ c = a^2 + b^2 \]

Explanation: this simply states that \( c \) equals the sum of the squares of \( a \) and \( b \). No manipulation is needed to obtain this form because the right-hand side is already expressed as a function of \( a \) and \( b \).
Solve for \( a \) (express \( a \) in terms of \( b \) and \( c \)).

Start from the original equation and isolate \( a^2 \) by subtracting \( b^2 \) from both sides:

\[ a^2 = c – b^2 \]

Now take square roots of both sides to solve for \( a \):

\[ a = \pm \sqrt{c – b^2} \]

Detailed explanation of each action:
- Subtracting \( b^2 \) from both sides uses the principle that equal quantities remain equal if the same number is subtracted from each side.
- Taking the square root of both sides gives two possibilities, positive and negative, because both \( +x \) and \( -x \) square to \( x^2 \). Hence the \( \pm \) symbol.
- Domain condition: for real values of \( a \), the quantity under the square root must be nonnegative. That is \( c – b^2 \ge 0 \). If \( c – b^2 < 0 \) and you restrict to real numbers, there is no real solution for \( a \); complex solutions exist and are given by the same formula with the complex square root.
Solve for \( b \) (express \( b \) in terms of \( a \) and \( c \)).

By symmetry, isolate \( b^2 \):

\[ b^2 = c – a^2 \]

Taking square roots:

\[ b = \pm \sqrt{c – a^2} \]

Explanation: the same reasoning as for \( a \) applies. For real \( b \), require \( c – a^2 \ge 0 \). If this fails, real solutions for \( b \) do not exist, but complex solutions do.
Implications and special cases.
- Nonnegativity of \( c \) for real \( a \) and \( b \).
  
  For any real numbers \( a \) and \( b \), squares are nonnegative, so their sum is nonnegative. Therefore if \( a \) and \( b \) are real, \( c \) must satisfy
  
  \[ c \ge 0 \]
- Case \( c = 0 \).
  
  If \( c = 0 \) and \( a \) and \( b \) are real, then both squares must be zero, so
  
  \[ a = 0 \quad \text{and} \quad b = 0 \]
- Given \( c \), infinitely many real pairs \( (a,b) \) when \( c > 0 \).
  
  If \( c \) is a fixed positive real number, there are infinitely many real pairs \( (a,b) \) satisfying the equation. For example, for any chosen \( a \) with \( |a| \le \sqrt{c} \), you can set
  
  \[ b = \pm \sqrt{c – a^2} \]
  
  and this yields a valid pair.
- Integer solutions.
  
  If \( a \), \( b \), \( c \) are required to be integers, then integer solutions occur when \( c \) equals a sum of two integer squares. There are classical results about which integers can be written as a sum of two squares, but describing those is a separate problem.

Summary (compact formulas)

\( c = a^2 + b^2 \)
\( a = \pm \sqrt{c – b^2} \) with the real-domain condition \( c – b^2 \ge 0 \)
\( b = \pm \sqrt{c – a^2} \) with the real-domain condition \( c – a^2 \ge 0 \)

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FAQs

What does \(a^2 + b^2 = c\) represent geometrically for real \(a,b,c\)?

For real numbers with \(c>0\), it's the circle centered at the origin with radius \(\sqrt{c}\). If \(c=0\) it's the origin point; if \(c<0\) there are no real solutions.

When do integers \(a,b\) exist for a given integer \(c\)?

An integer \(c\) is a sum of two integer squares iff in its prime factorization every prime \(p\equiv 3\pmod 4\) appears with an even exponent (classical sum-of-two-squares theorem).

Can a prime \(p\) be written as \(a^2+b^2\)?

Yes exactly when \(p=2\) or \(p\equiv 1\pmod 4\). Primes \(p\equiv 3\pmod 4\) cannot be expressed as a sum of two integer squares.

How do you solve for \(b\) given \(a\) and \(c\)?

Solve \(b=\pm\sqrt{\,c-a^2\,}\). For real \(b\) require \(c-a^2\ge 0\); for integer \(b\) require \(c-a^2\) to be a perfect square.

Is there a simple parametrization of all integer solutions \((a,b,c)\)?

There is no single simple parametrization for \(a^2+b^2=c\) with arbitrary integer \(c\). For fixed \(c\) you search integer pairs; for the related equation \(a^2+b^2=c^2\) there is a classical Pythagorean parametrization.

How many integer representations can a fixed \(c\) have?

What parity or modular restrictions exist for \(a^2+b^2\)?

Squares are \(0\) or \(1\pmod 4\), so \(a^2+b^2\) can be \(0,1,\) or \(2\pmod 4\). In particular it cannot be \(3\pmod 4\). If both \(a,b\) are odd then \(a^2+b^2\equiv 2\pmod 4\).

How do Gaussian integers help with \(a^2+b^2=c\)?

Write \(c=(a+bi)(a-bi)\) in the Gaussian integers. Unique factorization there translates the sum-of-two-squares problem into factoring primes and exponents, explaining existence and counting of representations.

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