1105 as Sum of Two Squares

Theorem
$1105$ can be expressed as the sum of two squares in more ways than any smaller integer:

Proof
We prove this result by computing the number of representations of $n$ as a sum of two squares for all numbers $n \le 1105$.

Let $a$ and $b$ be non-negative integers such that $a^2 + b^2 \le 1105$.

By the Derivative of Complex Polynomial, the derivative of the function $x \mapsto x^2$ is $x \mapsto 2 x$.

Since $2 > 0$ and $x \ge 0$ for any $x \in \hointr 0 \infty$, by Product of Real Numbers is Positive iff Numbers have Same Sign, for all $x \in \hointr 0 \infty$:


 * $2 x \ge 0$

By Real Function with Positive Derivative is Increasing, this shows that $x \mapsto x^2$ is increasing on the interval $\hointr 0 \infty$.

As $34^2 = 1156$ and $1156 > 1105$, it also follows that if $a \ge 34$:


 * $a^2 \ge 34^2 > 1105$

It also follows from the monotonicity of $x \mapsto x^2$ that since $b \ge 0$, it follows that $b^2 \ge 0$.

From Real Number Ordering is Compatible with Addition, we therefore have that for $a \ge 34$:


 * $a^2 + b^2 > a^2 > 1105$

To find all representations of numbers less than or equal to $1105$ as a sum of two squares, it therefore suffices to check how many times each number is represented as a sum of numbers of the form $a^2 + b^2$ where $b \le a$ and $a \le 33$.