Integer Surface Area

The smallest solution is $\{b,c,d\} = \{1,2,8\}$ , with an area of $A = \boxed{9}$ .

The area may be expressed as half of a cross product of vectors: $A = \tfrac12|\vec{BC}\times\vec{BD}| = \tfrac12\sqrt{(bc)^2 + (bd)^2 + (cd)^2}.$ In order for $(bc)^2 + (bd)^2 + (cd)^2$ to be a perfect square, they cannot all be odd. If they are all even, we can divide everything by two to find a smaller solution. (This may cause $A$ to be a half-integer, but we'll worry about that later.) We stipulate that $c > d$ have the same parity (both odd or both even). $b$ is the "odd duck" with the opposite parity.

Let $a = 2A$ . We rework the equation as $(b^2 + c^2)(b^2 + d^2) = a^2 + (b^2)^2.$ Note that with our choices of $b,c,d$ , the factors on the left are necessarily odd; $a$ has the same parity as $c$ and $d$ .

This brings us to the theory of (odd) sums of squares. If a sum of squares is a prime number, it is of the form $p = 4k+1$ ; conversely, any such prime may be written as a unique sums of squares. The product of sums of squares may always be written as a sum of squares. If a sum of squares is a compound number, there is more than one way to write it as a sum of squares. (The exception is a number of the form $p^2$ . If $p = x^2 + y^2$ then $p^2 = (x^2 - y^2)^2 + (2xy)^2$ is the only way to write $p$ as sum of squares.)

To save ourselves even more work, consider that if $b^2 + c^2$ and $b^2 + d^2$ are prime , the only ways to write the product as a sum of squares are $(b^2 + c^2)(b^2 + d^2) = (b(c \pm d))^2 + (b^2 \mp cd)^2.$ Comparing this to our earlier equation $(b^2 + c^2)(b^2 + d^2) = a^2 + (b^2)^2$ , we conclude that $b(c \pm d) = b^2;\ \ b^2 \mp cd = a.$ Thus we have $b = c \pm d$ , requiring $b$ to be even and $c,d$ to be odd. Also, $a = c^2 + d^2 \pm cd$ . The only glitch is that this makes $a$ odd, and therefore $A = a/2$ fractional. We resolve this by doubling $b,c,d$ (and thus quadrupling the area $A$ ). The area will always be even in this case.

Odd prime numbers of the form $4k + 1$ are as follows: $\begin{array}{cc} 5 & 2^2 + 1^2 \\ 13 & 2^2 + 3^2 \\ 29 & 2^2 + 5^2 \\ 53 & 2^2 + 7^2 \\ 17 & 4^2 + 1^2 \\ 41 & 4^2 + 5^2 \\ 97 & 4^2 + 9^2 \\ 37 & 6^2 + 1^2 \\ 61 & 6^2 + 5^2 \\ 73 & 8^2 + 3^2 \\ 89 & 8^2 + 5^2 \\ & \vdots \end{array}$

Thus we follow this recipe: choose an even value $b$ . In the table, find two corresponding odd values $c,d$ whose sum or difference is equal to $b$ . Select two squares $b^2 + c^2 \not = b^2 + d^2$ from the list. Calculate $a$ .

$\begin{array}{cccc} b & c,d & b^2 + d^2, b^2 + c^2 & a \\ \hline 2 & 1,3 & 5,13 & 7 \\ 2 & 3,5 & 13, 29 & 19 \\ 2 & 5,7 & 29, 53 & 39 \\ 4 & 1,5 & 17, 41 & 21 \\ 4 & 5,9 & 41, 97 & 61 \\ 6 & 1,5 & 37, 61 & 31 \\ 8 & 3,5 & 73, 89 & 49 \end{array}$

Recall that for a solution to the problem, we must double $a,b,c,d$ . Thus the smallest solution of this kind is $\{b,c,d\} = \{2,4,6\}$ with $A = 14$ .

However, it is also possible that if $b^2 + c^2$ and/or $b^2 + d^2$ is not prime . In that case, they will be a composite number, the product of two or more (odd) sums of squares. These composite numbers may be expressed as sums of squares in differrent ways. For instance,

$\begin{array}{ccc} 65 = 5\cdot 13 & 8^2 + 1^2 & 4^2 + 7^2 \\ 85 = 5\cdot 17 & 2^2 + 9^2 & 6^2 + 7^2 \\ 125 = 5\cdot 5^2 & 2^2 + 11^2 & 10^2 + 5^2 \\ 145 = 5\cdot 29 & 12^2 + 1^2 & 8^2 + 9^2 \\ 185 = 5\cdot 37 & 4^2 + 13^2 & 8^2 + 11^2 \\ 205 = 5\cdot 41 & 14^2 + 3^2 & 6^2 + 13^2 \\ 221 = 13\cdot 17 & 14^2 + 5^2 & 10^2 + 11^2 & \vdots \end{array}$

The smallest value of $(b^2+c^2)(b^2 + d^2)$ we can produce in this manner is $5\cdot 65 = (1^2 + 2^2)(1^2 + 8^2) = (1^2 + 2^2)(7^2 + 4^2) = (1\cdot 7 \pm 2\cdot 4)^2 + (1\cdot 4 \mp 2\cdot 7)^2 = \begin{cases} 15^2 + 10^2 \\ 1^2 + 18^2 \end{cases}$ Here we have $b = 1$ , so we look for an expression of the form $a^2 + (b^2)^2 = a^2 + 1^2$ . Clearly, this happens for $a = 18$ , so that $(1^2 + 2^2)(1^2 + 8^2) = 18^2 + (1^2)^2,\ \ \ \ \{b,c,d\} = \{1,2,8\},\ \ \ A = \boxed{9}.$ This is an improvement over the earlier solution $A = 14$ (only because the fractional value $A = 3\tfrac12$ was not allowed). It is clear that choosing greater factors will lead to greater values of $b,c,d$ and therefore greater areas. For instance, another relatively small solution is $13\cdot 85 = (2^2 + 3^2)(2^2 + 9^2) = (2^2 + 3^2)(6^2 + 7^2) = 33^2 + (2^2)^2;\ \ \ \text{doubled to}\ \{b,c,d\} = \{4,6,18\}, A = 66.$

Denote $BC=\sqrt{b^2+c^2},\,BD=\sqrt{b^2+d^2},\,CD=\sqrt{c^2+d^2}$ , $s=\dfrac{BC+BD+CD}2$ , and apply Heron's formula $P_{\Delta\,BDC}=\sqrt{s(s-BC)(s-BD)(s-CD)}=\dfrac12\sqrt{b^2c^2+b^2d^2+c^2d^2}$ , and we find the minimal integer value of $\dfrac12\sqrt{b^2c^2+b^2d^2+c^2d^2}$ in distinct integers $b,c,d$ . Since these number are to be positive, and we are seeking the minimum of that expression, we set the least value, for instance, for $d$ , i.e. $d=1$ . Since these numbers are distinct, we choose the least value, for instance, for $c$ , i.e. $c=2$ . Thus, the expression is reduced to $\dfrac12\sqrt{5b^2+4}=\sqrt{5\Big(\dfrac b2\Big)^2+1}$ , and that means $b$ must be even. Denote $b=2k$ , and we are looking for the first perfect square $5k^2+1$ . The immediate check gives rise to $k=4$ , meaning that $9$ is the minimal value, reached for $d=1,\,c=2,\,b=8$ .