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Math Help - Prove that x^2 - y^2 = 102 has not integer solutions

  1. #1
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    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?

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  2. #2
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    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 $x^2 - y^2 = 102 \implies (x - y)(x + y) = 102.$

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

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

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

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

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

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

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

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

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

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

    $\therefore y \not \in \mathbb Z.$
    Last edited by JeffM; June 10th 2014 at 06:55 PM. Reason: typos
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  3. #3
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    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...
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    Re: Prove that x^2 - y^2 = 102 has not integer solutions

    Hello, ShadowKnight8702!

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

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

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

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

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