Difference of Squares (3)

The numbers $N$ for which it is true are $15,21,24,27,32,33,35,39,40,51,55,56,57,60,64,65,69,77,81,84,85,87,88,91,93,95.$

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Basically, for every factoring $N = xy$ ( $x < y$ ) we have $N = s^2 - d^2$ with $s = \tfrac12(x + y)$ and $d = \tfrac12(y - x)$ . Thus, the number of such pairs $(s,d)$ is half of the number of factors of $N$ .

Also, the number of positive factors of $N = 2^a\cdot 3^b \cdot 5^c\cdots$ is $\delta(N) = (a+1)(b+1)(c+1)\cdots$ .

There are some exceptions to consider:

• If $x$ and $y$ have opposite parity (one odd, one even), the values of $s$ and $d$ are not integers. Therefore, if $N$ is even, we limit ourselves to factorings into two even numbers.

• If $N$ is a square, the factoring $N = x^2$ would give $N = x^2 - 0^2$ , and must therefore ruled out.

With this information, we tackle the possible cases.

Case Ia : None of the exceptions come into play; $N$ is odd and not a square. Let $N = 3^b \cdot 5^c\cdot 7^d\cdots$ , then we require $\tfrac12(b+1)(c+1)(d+1)\cdots = 2\ \ \ \ \text{i.e.}\ \ \ \ (b+1)(c+1)(d+1)\cdots = 4.$ This can be accomplished in two ways:

• Precisely one of the exponents $b,c,d,\dots$ is equal to 3. This means that $N$ is the cube of an odd prime number.

• Precisely two of the exponents $b,c,d,\dots$ are equal to 1. This means that $N$ is the product of two district odd primes .

Case Ib : $N$ is an odd square. Let $N = 3^{2\beta}\cdot 5^{2\gamma}\cdot 7^{2\delta}\cdots$ , then we find (after ruling out the factor $\sqrt N$ ): $(2\beta+1)(2\gamma+1)(2\delta+1)\cdots = 5.$ This requires that one of $2\beta,2\gamma,2\delta,\dots$ is 4; in other words, $N$ is the fourth power of an odd prime .

Case II : $N$ is even. Since we require the factors $x,y$ to be even, we might as well work with $N' = x'y'$ , where $N' = N/4$ , $x' = x/2$ and $y' = y/2$ . Thus we are back to case I, except now we allow 2 as a prime factor and multiply each number we find by 4. Thus:

• $N$ is four times the cube of a prime number (including $4\cdot 2^3$ ).
• $N$ is four times the product of two distinct primes (including $4\cdot 2q$ ).
• $N$ is four times the fourth power of a prime number (including $4\cdot 2^4$ ).

$\begin{array}{ccl} \text{case} & \text{form} & \text{values of}\ N\ \text{up to 100} \\ \hline \text{Ia(i)} & p^3 & 27 \\ \text{Ia(ii)} & pq & 15,21,33,39,51,57,69,87,93 \\ & & 35,55,65,85,95 \\ & & 77,91 \\ \text{Ib} & p^4 & 81 \\ \text{IIa(i)} & 4\cdot 2^3 & 32 \\ & 4\cdot p^3 & - \\ \text{IIa(ii)} & 4\cdot 2q & 24,40,56,88 \\ & 4\cdot pq & 60, 84 \\ \text{IIb} & 4\cdot 2^4 & 64 \\ & 4\cdot p^4 & - \\ \hline \end{array}$

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A complete list of the 26 solutions and the two ways to write each as a difference of squares:

$\begin{array}{rcc} N & & \\ \hline 15 & 8^2 - 7^2 & 4^2 - 1^2 \\ 21 & 11^2 - 10^2 & 5^2 - 2^2 \\ 24 & 7^2 - 5^2 & 5^2 - 1^2 \\ 27 & 14^2 - 13^2 & 6^2 - 3^2 \\ 32 & 9^2 - 7^2 & 6^2 - 2^2 \\ 33 & 17^2 - 16^2 & 7^2 - 4^2 \\ 35 & 18^2 - 17^2 & 6^2 - 1^2 \\ 39 & 20^2 - 19^2 & 8^2 - 5^2 \\ 40 & 11^2 - 9^2 & 7^2 - 3^2 \\ 51 & 26^2 - 25^2 & 10^2 - 7^2 \\ 55 & 28^2 - 27^2 & 8^2 - 3^2 \\ 56 & 15^2 - 13^2 & 9^2 - 5^2 \\ 57 & 29^2 - 28^2 & 11^2 - 8^2 \\ 60 & 16^2 - 14^2 & 8^2 - 2^2 \\ 64 & 17^2 - 15^2 & 10^2 - 6^2 \\ 65 & 33^2 - 32^2 & 9^2 - 4^2 \\ 69 & 35^2 - 34^2 & 13^2 - 10^2 \\ 77 & 39^2 - 38^2 & 9^2 - 2^2 \\ 81 & 41^2 - 40^2 & 15^2 - 12^2 \\ 84 & 22^2 - 20^2 & 10^2 - 4^2 \\ 85 & 43^2 - 42^2 & 11^2 - 6^2 \\ 87 & 44^2 - 43^2 & 16^2 - 13^2 \\ 88 & 23^2 - 21^2 & 13^2 - 9^2 \\ 91 & 46^2 - 45^2 & 10^2 - 3^2 \\ 93 & 47^2 - 46^2 & 17^2 - 14^2 \\ 95 & 48^2 - 47^2 & 12^2 - 7^2 \\ \hline \end{array}$

Mathematica

Count[Array[Length@Solve[a^2-b^2==#&&0<b<a,{a,b},Integers]&,100],2]

26