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The question is :

Write 1009 in the form $(x^2+a\,y^2)$ for $a=1,2,3$. Where $x$ and $y$ are integers.

It is a question where we have to show our work but I know no method other than trial and error which is obviously not allowed... So please tell the correct method/trick to solve the question. Although I found answer for the first part by guess work which is: $(28)^2 + (15)^2 =1009$. But as I mentioned above we can not simply write this we have to show the work so please help.

A.G.
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Akshaj Bansal
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4 Answers4

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One way is check if there is an element in $\mathbb{Z}[\sqrt D ]$ with $D$ square-free (i.e if the form $a+b\sqrt{D}$) with norm 1009.

That number $z=a+b\sqrt{D}$ must to be prime in $\mathbb{Z}[\sqrt D ]$ because divisibility of its norm.

$15^2+28^2=1009$

$19^2+2\cdot18^2=1009$

$31^2+3\cdot 4^2=1009$

$15^2+4\cdot 14^2=1009 $

$17^2+5\cdot 12^2=1009 $

  • Yeah, and how did you calculate all those values? – nonuser Jul 09 '19 at 14:46
  • the first value was given by OP. the third, fourth, and last by calculator. But the second through Gauss Representation of cuadratic forms. OP would like to check https://math.stackexchange.com/a/236147/681693 after a reading of https://www.amazon.com/dp/1118390180/?tag=stackoverfl08-20 –  Jul 09 '19 at 14:54
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COMMENT.-Theoretically, this problem involves the representation of a rational integer by a binary quadratic form $ax^2+by^2=n$ . In this case for a prime $n=1009$ (when $a=b=1$ one has the well known Fermat's theorem and $x^2+y^2=1009$ has solution because $1009\equiv 1\pmod4$ but, for example, for the prime n = 1019 there is no solution ).

For the other two questions the conditions of solvability are fulfilled too but these ones are something more elaborate.These conditions can be found in the book Diophantine equations,by J.L. Mordell page 164-166.

When you know that the equations have a solution you can try to find them, as long as the number $n$ is not very large.

Piquito
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You can solve this easily using Excel. Clearly $0\leq x\leq \lfloor \sqrt{1009}\rfloor=31$ and $$ 1009=x^2+a y^2 \rightarrow y=\sqrt{1009-x^2\over a} $$ (here $a=1,2,3$). Just build a table of $y=\sqrt{1009-x^2\over a}$ for the 32 possible values of $x$ and check for integer $y$'s. You get the solutions $$ 1009= 28^2+15^2 = 19^2 + 2\times 18^2 = 31^2 + 3 \times 4^2. $$ enter image description here

Of course you can also easily program this if you know how to code.

A.G.
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    If trial and error is not allowed surely an excel program isn't either. – fleablood Jul 09 '19 at 16:53
  • It was a subjective mathematics paper where obviously you can not use excel . You were required to provide a solution which involves mathematical theorems , working and formulas ,although thanks for your answer – Akshaj Bansal Jul 09 '19 at 17:29
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$$1009 = 15^2+28^2$$

$$1009 = 19^2+2\cdot 18^2$$

$$1009 = 31^2+3\cdot 4^2$$

Well I calulate this values manualy with the use of calculator. Just write $y = \sqrt{1009-x^2}$ for the first case where $x\leq 31 =[\sqrt{1009}]$.

$y = \sqrt{1009-x^2 \over 2}$ for the second case where $x\leq 31$ and odd.

$y = \sqrt{1009-x^2 \over 3}$ for the last case where $x\leq 31$.

nonuser
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  • Wolfram gives these solutions but without explanation. VERY IMPORTANT (for me): I DID NOT PUT THE DOWNVOTE! – Piquito Jul 09 '19 at 14:35
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    Downvote here, because no explanation – ESCM Jul 09 '19 at 14:36
  • @EduardoS. Any better now? – nonuser Jul 09 '19 at 14:43
  • Any better now? @Piquito – nonuser Jul 09 '19 at 14:47
  • Dear friend: I'm sorry but I think about what you would have done if instead of 1009 it would have been a 40-digit number (I admit that I'm not very flexible in this). Regards – Piquito Jul 09 '19 at 15:49
  • On a certain occasion, I took an advanced course in differential equations. To my surprise throughout the semester, not a single differential equation was solved. Only theorems of existence of solutions interested in that advanced perspective. – Piquito Jul 09 '19 at 16:10