Find all pairs of positive integers whose squares differ by 400?

**Rinnie** · October 25th 2008, 11:34 AM

I have pairs, but I need to find an organzied method for finding them?

**Peritus** · October 25th 2008, 02:47 PM

$<br /> m^2 - n^2 = 400\quad m,n \in Z^ +$

$\left( {m - n} \right)\left( {m + n} \right) = 400$

thus m-n, or m+n must equal one of the divisors of 400, to satisfy the equation, and so that m,n will be positive integers.

for example, 10 is a divisor of 400:

m - n = 10
m + n = 40

-> m = 25, n = 15

**Soroban** · October 25th 2008, 03:19 PM

Hello, Rinnie!

Find all pairs of positive integers whose squares differ by 400.

Let $x,y$ be positive integers such that: . $x^2-y^2 \:=\:400$

Then we have: . $(x+y)(x-y) \:=\:P\cdot Q$ ... where $P\cdot Q \:=\:400$

Solve the system: . $\begin{array}{ccc}x + y &=& P \\ x-y&=& Q\end{array}\quad\Rightarrow\quad\begin{Bmatrix}x &=&\frac{P+Q}{2} \\ \\[-4mm] y &=&\frac{P-Q}{2} \end{Bmatrix}$

Since $x\text{ and }y$ are integers, $P\text{ and }Q$ have the same parity;
. . both even or both odd.

The only ways to factor 400 into two factors with the same parity are:
. . $(200, 2),\;\;(100,4,),\;\;(50,8),\;\;(40,10),\;\;(20,20)$

Then we have:

. . $\begin{array}{c|cc}<br /> (P,Q) & (x,y) \\ \hline (200,2) & (101,99) \\ (100,4) & (52,48) \\ (50,8) & (29,21) \\ (40,10) & (25,15) \\ (20,20) & {\color{red}\rlap{/////}}(20,0) & {\color{red}\text{not positive}}\end{array}$

Therefore: . $101^2-99^2 \:=\:52^2-48^2 \:=\:29^2-21^2 \:=\:25^2-15^2 \:=\:400$

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