# Thread: Prove that x^2 - y^2 = 102 has not integer solutions

1. ## Prove that x^2 - y^2 = 102 has not integer solutions

Hi everyone,

I came across this question after reading up on integer properties. I have absolutely no idea where to start. I tried manipulating the equation into different forms of difference of two squares. I also tried doing several modulo operations. Nothing I did seemed to work. How would I go about solving this problem?

Thanks

2. ## Re: Prove that x^2 - y^2 = 102 has not integer solutions

Maybe start by finding all pairs of integer factors of 102.

$\pm 1\ and\ \pm 102,\ \pm 2\ and\ \pm 51,\ \pm 3\ and\ \pm 34,\ or\ \pm 6\ and\ \pm 17.$

As you saw $x^2 - y^2 = 102 \implies (x - y)(x + y) = 102.$

$x\ and\ y\ are\ integers \implies (x - y) = m\ and\ (x + y) = n\ are\ integers.$

$x - y = m\ and\ x + y = n \implies x - y + x + y = m + n \implies x = \dfrac{m + n}{2} \implies |x| = \bigg |\dfrac{m + n}{2} \bigg|.$

$So\ x \in \mathbb Z\ only\ if\ |m + n|\ is\ even.$

$But\ mn = 102 \implies Signum(m) = Signum(n) \implies |m + n| = |m| + |n|.$

$So\ x \in \mathbb Z\ only\ if\ |m| + |n|\ is\ even.$

$|1| + |102| = 103,\ not\ even.$

$|2| + |51| = 53,\ not\ even.$

$|3| + |34| = 37,\ not\ even.$

$|6| + |17| = 23,\ not\ even.$

$So\ x \not \in \mathbb Z.$

$\therefore y \not \in \mathbb Z.$

3. ## Re: Prove that x^2 - y^2 = 102 has not integer solutions

The only integer factors of 102 are 1x102, 2x51, 3x34, 6x17. As you have said, x^2 - y^2 = (x - y)(x + y).

Let's try using 1 and 102. That would give two equations x - y = 1 and x + y = 102. Solving them simultaneously gets x = 103/2 and y = -101/2, which are not integers.

See what you get when you try using the other factor combinations...

4. ## Re: Prove that x^2 - y^2 = 102 has not integer solutions

$\displaystyle \text{Show that }\,x^2-y^2 \:=\:102\,\text{ has no integer solutions.}$

We have: .$\displaystyle (x+y)(x-y) \:=\:p\!\cdot\!q$
. . where $\displaystyle p,q$ are integers, $\displaystyle p > q,$ and $\displaystyle p\!\cdot\!q = 102.$

We have: .$\displaystyle \begin{Bmatrix}x+y \;=\;p \\ x-y \;=\;q \end{Bmatrix}$

Solve the system: .$\displaystyle \begin{Bmatrix}x \;=\;\frac{p+q}{2} \\ y \;=\;\frac{p-q}{2}\end{Bmatrix}$

Since $\displaystyle x$ and $\displaystyle y$ are integers, we see that $\displaystyle p$ and $\displaystyle q$ must have the same parity
. . (both are even or both are odd).
But these are the only factorizations of 102: .$\displaystyle \begin{bmatrix} 1\!\cdot\!102 \\ 2\cdot 51 \\ 3\cdot 34 \\ 6\cdot17 \end{bmatrix}$
Therefore, there are no integers solutions.

5. ## Re: Prove that x^2 - y^2 = 102 has not integer solutions

Perhaps just a tiny bit faster: 102= 2(3)(17) has only a single factor of 2. That means that if pq= 102 one of p, q must be even and the other odd. But, as Soroban showed, the two factors, x+ y and x- y, can be written as (p+ q)/2 and (p- q)/2. Those are not integer.

,
,

,

# x^2 102 = y^2

Click on a term to search for related topics.