If two positive integers m and n, both bigger than 1, satisfy the equation 2005^2 + m^2 = 2004^2 + n^2 , find the value of m + n – 200

Question

I tried to use the difference of two squares and was able to get product of two quantities on either side of the equation ,but i am stuck there .

Answer 1

From your equation, we have: $$n^2-m^2=2005^2-2004^2$$ It can be shown also that $A^2-B^2=(A+B)(A-B)$, and so: $$(n-m)(n+m)=(2005-2004)(2005+2004)$$ This leads to: $$n-m=1 \land n+m=4009$$ And so: $$n=2004 \land m=2005$$ This result could be obtained also, and more simply, observing the equation above. So, one possible value asked is: $$m+n-200=2004+2005-200=3809$$ Another solution is given by: $$n-m=19 \land n+m=211$$ And so: $n=115 \land m=96$. This solution comes from the factorization of $4009$ into $19\cdot 211$.

Answer 2

We have $$ 19\cdot 211=4009=2005^2-2004^2=n^2-m^2=(n+m)(n-m). $$ One solution is $m+n=211$ and $m-n=19$, so that $(m, n)=(96,115)$. The other one is for you to find out.

Answer 3

The equation $$2005^2 + m^2 = 2004^2 + n^2$$ gives: $$2005^2 - 2004^2 = n^2 - m^2,$$ $$(2005 - 2004)(2005 + 2004) = (n-m)(n+m),$$ $$(n-m)(n+m) = 4009.$$ Factoring $4009$ we find $4009=19\cdot 211$, so it can be either $$\begin{cases}n-m=19\\n+m=211\end{cases}$$ or $$\begin{cases}n-m=1\\n+m=4009\end{cases}.$$ And the solutions are, respectively,

$$\begin{cases}m=96\\n=115\end{cases}$$

or

$$\begin{cases}m=2004\\n=2005\end{cases}.$$

From here you can calculate $m+n–200$ in both cases.

Answer 4

Difference of two squares give me

$2005^2 + m^2 = 2004^2 + n^2$

$2005^2 - 2004^2 = n^2 - m^2$

$(2005 - 2004)(2005+2004) = (n-m)(n+m)$

$4009 = (n-m)(n+m)$.

$n+m > 0$ so $n-m > 0$ and $n > m$ and $0 < n-m < n+m$.

Google search to find the prime factors of $4009$ and $4009 = 19*211$.

So there are two possibilities.

$n-m = 1$ and $n+m = 4009$ and $n+m -200 = 3809$. (This would make $n=2005$ and $m=2004$. I guess it should always be obvious that one solution to $J^2 + n^2 = K^2 + m^2$ is $m =J; n=K$....)

Or $n-m = 19$ and $n+m = 211$ and $n+m-200 = 11$. (This would make $n=115$ and $m = 96$.)

Was there something in the question that $\{n,m\}\ne \{2005, 2004\}$? Or I wonder of the authors just assumed $4009$ was prime (weird assumption... not all odd numbers are prime, of course). Or maybe the author wanted you to list all possible answers.