Given the equation: $x^2+5xy=5775$.
How can i find the positive integer solutions of it?
I can write it as $x(x+5y)=5775$ but it seems to be hard to guess.
Thanks.
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well, $3,5,7,11$? – david Oct 29 '14 at 09:26
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1Reducing the original equation mod $5$ shows that $x \equiv 0 \bmod 5$. Does this narrow down the possibilities for you? – fretty Oct 29 '14 at 09:26
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1@david I think there are many more divisors. – fretty Oct 29 '14 at 09:27
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@fretty - sorry but can't see how. – david Oct 29 '14 at 09:38
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Ok well maybe I can write it in an elementary way. Rearranging gives $x^2 = 5775 - 5xy = 5(1155 - xy)$. Can you now see that $x$ has to be a multiple of $5$? – fretty Oct 29 '14 at 09:53
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http://math.stackexchange.com/questions/991773/solve-this-simple-equationax2byxc-0 – Bumblebee Oct 29 '14 at 11:35
3 Answers
1
$$5775=3*5*5*7*11$$ Using Fretties comment we have $x=5, 15, 35,55. $ For example $x=5,y+1=407.$
But if $x=105$ then $ 105+5y>\frac{5775}{105} $
Liza
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Mod 5, you can see $x\equiv0$. So $x=5k$, and now $$25k^2+25ky=5775$$ $$k^2+ky=231$$ $$k(k+y)=231$$ So $k$ is a factor of $231=3\cdot7\cdot11$. Since $y$ is positive, $k$ must be smaller than its cofactor $k+y$. So there are $\frac122^3=4$ options for $k$ (explicitly they are $1,3,7,11$), and then $y$ is determined as $\frac{231}{k}-k$ (and $x=5k$.)
2'5 9'2
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$x^2=5(1155-xy)$
x divided into 5 ⇨ x = 5k
$25k^2=5(1155-xy) ⇨ 5k^2 = 1155-xy ⇨ ky=231-k^2 ⇨ y={231\over k}-k$
231 divided into k.
231 dividers:
1, 3, 7, 11
21, 33, 77, 231
Check all $k<[\sqrt{231}]=15:$
for k=1 ⇨ x=5
Using $ky=231-k^2 ⇨ y=230$
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for $k>15 ⇨ y<0$
It does not suit us. Solutions:
$x=5,y=230; x=15, y=74; x= 35, y=26; x=55, y=10.$
Art Fray
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