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I am relatively new at solving these kind of equation and was wondering if someone can help with a step by step for an odd formed Diophantine equation. The particular equation I am trying to solve is $-x^2 -2xy + 44y = 43$.

The integer solutions are not in general form but are

$${y = -286, x = 23}$$ $${y = -286, x = 549}$$ $${y = -46, x = 39}$$ $${y = -46, x = 53}$$ $${y = 2, x = 5}$$ $${y = 2, x = -9}$$ $${y = 242, x = -505}$$ $${y = 242, x = 21}$$

Ryan Topps
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2 Answers2

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Hint: The equation is equivalent to $$ (x+2y+22)(22-x)=17\cdot31 $$

robjohn
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  • Mathematica: {#,(43+#^2)/(44-2#)}&/@(22-Flatten[Outer[Times,{1,-1},{1,17},{1,31}]]) – robjohn Sep 29 '18 at 15:59
  • Thanks, question though is there a way to solve this not using the prime factorization but rather with just modulo math? – Ryan Topps Sep 29 '18 at 17:14
  • Good luck with that since $22-x$ must divide $527$. If that is really what you want, you might want to add that requirement to your question. – robjohn Sep 29 '18 at 17:19
  • I just was wondering as if the numbers get larger the prime factorization is hard to get. Thanks for pointing me in that direction though. – Ryan Topps Sep 29 '18 at 17:26
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When you have a quadratic, the quadratic formula is your friend. Write it as $$x^2+2xy+43-44y=0\\x=\frac 12\left(-{2y}\pm\sqrt{4y^2-172+176y}\right)\\ =\frac 12\left(-{2y}\pm\sqrt{(2y+44)^2-1764}\right)$$ and you need to look for $y$s that will make the square root an integer. If you define $z=2y+44$ and require $z$ to be even you are looking at $$z^2-42^2=w^2\\(z+w)(z-w)=42^2$$ and look at the factorizations of $42^2$ into even numbers.

Ross Millikan
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