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$$k*100x^2+20x=100y^2+20y$$ I have the following equation where $k $ and $y $ and $x $ are all positive intergers. Also $k\gt1$ and $k\neq x\neq y $ Is there any solutions to this equation?

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    $\begingroup$ Sure $$x = y = 0$$ Also, $$k = 1, x = y$$ $\endgroup$
    – Moo
    May 31, 2018 at 21:55
  • $\begingroup$ I forgot to say positive $\endgroup$ May 31, 2018 at 21:55
  • $\begingroup$ @moo see my edited response $\endgroup$ May 31, 2018 at 21:58
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    $\begingroup$ Well $k = 37; x = 1; y=6$ is a solution. I have found that for there to be a solution we must have $y = x + 5j$. And so $(k-1)x^2 = 10xj + 25j^2 + j$ but... If we set $x = 1$ we can set $j$ to any integer value. But I don't see how to solve in general. $\endgroup$
    – fleablood
    May 31, 2018 at 23:24

3 Answers 3

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Yes. The MATLAB code below

f_L = @(x, k) k * 100 * x * x + 20 * x;
f_R = @(y) 100 * y * y + 20 * y;

LIMIT = 100;
i = 1;

for k = 2:1:LIMIT
    for x = 1:1:LIMIT
        for y = 1:1:LIMIT
            if k ~= x && k ~= y && x ~= y
                if f_L(x, k) == f_R(y)
                    arr(i, 1) = k;
                    arr(i, 2) = x;
                    arr(i, 3) = y;
                    i = i + 1;
                end
            end
        end
    end
end

produces $$\begin{array}{|c|c|c|} \hline k & x & y \\\hline 19 & 3 & 13 \\ 21 & 7 & 32 \\ 28 & 14 & 74 \\ 29 & 8 & 43 \\ 37 & 1 & 6 \\ 59 & 6 & 46 \\ 73 & 2 & 17 \\\hline \end{array}$$

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Final answer. For any $x$ solve (if possible) $10xj + 25j^2 + j\equiv 0 \mod x^2$.

Then $y = x + 5(j + mx^2)$ and $k = \frac {10x(j+mx^2) + 25(j+mx^2)^2 + (j+mx^2)}{x^2} + 1$ will be solutions.

Why? Well see my working below.

====== working below (or old answer) =====

Divide everything by $20$.

So $k*5x^2 + x = 5y^2 + y$

So $x - y = 5y^2 - k5x^2$. Assuming $x\ne y$ then

$x-y = 5(y^2 - x^2) - (k-1)5x^2$

$1 = 5(y + x) - 5\frac {(k-1)x^2}{x-y}$

So $x-y|5(k-1)x^2$ but assuming $k > 1$ then $x-y \not \mid (k-1)x^2$ so, as $5$ is prime $5|x-y$ and $x \equiv y \mod 5$.

So $y = x + 5j$ for some positive $j$.

$k*5x^2 + x = 5(x + 5j)^2 + x + 5j$

$k*5x^2 = 5x^2 + 50xj + 125j^2 + 5j$

$(k-1)x^2 = 10xj + 25j^2 + j$

Welp... if we set $x=1$ we can get an infinite number of solutions.

$k = 10j + 25j^2 + j + 1$

So $x = 1$ and $k=37;y=6$ and $k= 123; y = 11$ etc. will be solutions.

.... otherwise....

Let $x = x$ (as Laurie Anderson used to say)

Then solve for $10j + 15j^2 + j \equiv 0 \mod x^2$ if possible.

Let $k - 1 = \frac {10j + 15j^2 + j}{x^2}$. That Will give us solutions.

If $x = 2$ then $j$ can be any multiple of $4$ of $3 \mod 4$.

And if $x = 3$ we need $(k-1)9 = 31j + 25j^2$ and ...$31j + 25j^2 \equiv 4j-2j^2\mod 9$ so ... $j = 2+ 9m$ is a solution (as are $j = 9m$).

$x = 3; y = 13; k = \frac {162}9 + 1= 19$.

Ex: $x = 2$ then $4(k-1)= 21j + 25j^2$ so $21j + 25j^2 \equiv j + j^2 \equiv 0 \mod 4$ and so

$j = 3$

$x = 2; y=17; k= \frac {63 + 225}4 + 1 = 73$

.....

$37*100*1^2 + 20*1 = 100*6^2 + 6*20$

$3720 = 3600 + 120$. Check

$123*100*1^2 + 20 = 100*11^2 + 11*20$

$12320 = 12100 + 220$. Check.

$19*100*3^2 + 20*3 = 100*13^2 + 13*20$

$19*9*100 + 60 = 16900 + 260$

$17160 = 17160$. Check.

$x = 2; y = 17;k=73$ and

$73*100*2^2 + 2*20 = 100*17^2 + 17*20$

$29240 = 28900 + 340$. Check.

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  • $\begingroup$ thanks for your answer! $\endgroup$ May 31, 2018 at 23:38
  • $\begingroup$ (All variables are positive intergers) The original problem was proving that $100x^2+20x\neq m! $ for all values of $x$ and $m$. Stupidly I thought someone like this. $\endgroup$ Jun 1, 2018 at 0:07
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    $\begingroup$ $100x^2 + 20x =x*4*5(5x + 1)$ Consider the prime factorization of $x$ and of $m!$. $\endgroup$
    – fleablood
    Jun 1, 2018 at 0:27
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    $\begingroup$ $x$ and $5x+1$ are relatively prime. Let $p$ and $q$ be primes larger than $x$ but less than $m$ then $p$ and $q$ divide $5x + 1$ but if $x > 5$ then $5x + 1 < pq$. $\endgroup$
    – fleablood
    Jun 1, 2018 at 0:36
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    $\begingroup$ No, I meant less than $m$. $m!$ must be divisible by every number less than and including $m$. The larger $m$ is the more incrediably unlikely the factors $x, 4,5$ and $5x +1$ are going to cut it. If $p, q$ are prime and $x < p < m$ and $x < q < m$ then $pq|5x + 1$ and if $p,q > 5$ that is impossible. It'll take a small amount of parameter setting to show that for most $m$ there will have to be such primes and to make exceptions for the very small values of $m$ where this isn't true but that's a solid game plane. $\endgroup$
    – fleablood
    Jun 1, 2018 at 4:49
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Definitely. To do this easily, simply set x=0, making the value of k irrelevant. So you now have $100y^2+20y=0$. This can obviously be simplified to $5y^2+y=0$. Factoring out y, we get $y(5y+1)=0$, making $y=-\frac{1}{5}$ and $y=0$ solution. Since $y\neq x$, $y=-\frac{1}{5}$

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  • $\begingroup$ this is a nice solution however are all the variables are positive intergers. $\endgroup$ May 31, 2018 at 22:15

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