# to prove something is a square of rational

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#1
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#2
The first part is very obvious, but I have no idea how to do the second part
help!!!
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2 years ago
#3
There may be a less laborious method, but have you tried writing and . Multiplying these together, we obtain . Doing the same for and subtracting the two, we obtain which is beginning to look promising.
Last edited by max271; 2 years ago
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2 years ago
#4
Not done it fully but was trying to square up the first expression. It has some of the terms that are needed but haven't fully simplified.
The xy terms for the expression are x^4y^4 which is the square of the first expression and Im presuming that you can use a similar squaring argument.

(Original post by cxs)
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2 years ago
#5
If it's any consolation, it's probably from Hardy's Pure Mathematics - reprinted from the 1899 Maths. Tripos!
Last edited by ghostwalker; 2 years ago
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2 years ago
#6
(Original post by cxs)
Hmmm the thing is the second reminds me very much of the discriminant b^2-4ac, so perhaps if you manage to combine both equations in terms of one variable (like x, y or xy) as a quadratic, you can deduce that the discriminant must be the square of a rational for otherwise, the roots will be irrational which cannot be the case as you are given that x and y are rational.

(Original post by ghostwalker)
If it's an consolation, it's probably from Hardy's Pure Mathematics - reprinted from the 1899 Maths. Tripos!
O wow I was thinking this was from a recent A level paper. Good to know im not the only one who finds this hard hehe
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2 years ago
#7
(Original post by Alpacamatrix)
Hmmm the thing is the second reminds me very much of the discriminant b^2-4ac, so perhaps if you manage to combine both equations in terms of one variable (like x, y or xy) as a quadratic, you can deduce that the discriminant must be the square of a rational for otherwise, the roots will be irrational which cannot be the case as you are given that x and y are rational.
That was the methodology (at least the one I used) for the first expression. And the form of the second does seem to point in that direction too, though getting the right equation to use the discriminant on is the \$64,000 question. Not followed through the previous suggestions though.
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#8
Yes! The book is great, except,
(Original post by ghostwalker)
If it's any consolation, it's probably from Hardy's Pure Mathematics - reprinted from the 1899 Maths. Tripos!
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2 years ago
#9
some websites compile the answers online in a PDF which you could search for
(Original post by cxs)
Yes! The book is great, except,
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#10
I tried this but, how to go further???
Feel like I lost some key points.
(Original post by max271)
There may be a less laborious method, but have you tried writing and . Multiplying these together, we obtain . Doing the same for and subtracting the two, we obtain which is beginning to look promising.
My first thought is to eliminate something, like we can express xy, x^2,.. in two ways and get

But I do not think I can continue in this way...
Maybe I should derive functions about x^4y^4
(Original post by mqb2766)
Not done it fully but was trying to square up the first expression. It has some of the terms that are needed but haven't fully simplified.
The xy terms for the expression are x^4y^4 which is the square of the first expression and Im presuming that you can use a similar squaring argument.
Last edited by cxs; 2 years ago
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#11
??!!!!
Perhaps I just did a rough search weeks ago... I should search again
(Original post by Satyr)
some websites compile the answers online in a PDF which you could search for
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2 years ago
#12
The first part was about x^2y^2 and the terms in the second expression correspond to x^4y^4, but doing a naive squaring gives too many terms which need to be squared so some form of factorisation would be needed if this is the right approach. Currently scratching my head ...

(Original post by cxs)
I tried this but, how to go further???
Feel like I lost some key points.

My first thought is to eliminate something, like we express xy, x^2,.. in two ways and get

But I do not think I can continue by this way...
Maybe I should derive functions about x^4y^4
0
2 years ago
#13
(Original post by cxs)
??!!!!
Perhaps I just did a rough search weeks ago... I should search again
Good luck. I've tried a couple of searches, but to no avail. I did come across a solution to the previous question on stackexchange, but nothing specific to this problem nor an overall set of solutions.

I've now looked at max271's method - cudos to them for slogging through the algebra - and whilst agreeing so far, the x^2 term and y^2 term, if we could get rid of them, we'd be home and dry, but unfortunately, I can't see a way to do that.
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#14
I did not manage finding the answers to the whole book. Satyr
But I find this!!!
https://math.stackexchange.com/quest...=SearchResults
It is solved in a really brilliant way.
(Original post by ghostwalker)
Good luck. I've tried a couple of searches, but to no avail. I did come across a solution to the previous question on stackexchange, but nothing specific to this problem nor an overall set of solutions.

I've now looked at max271's method - cudos to them for slogging through the algebra - and whilst agreeing so far, the x^2 term and y^2 term, if we could get rid of them, we'd be home and dry, but unfortunately, I can't see a way to do that.
Last edited by cxs; 2 years ago
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#15
It turns out to be a really hard question.
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#16
Post-mortem: Looking at this question again it's not that hard
all we need to do is to find a equation satisfied by y.
Last edited by cxs; 2 years ago
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2 years ago
#17
(Original post by cxs)
Post-mortem: Looking at this question again it's not that hard
all we need to do is to find a equation satisfied by y.
Unparseable or potentially dangerous latex formula. Error 5: Image dimensions are out of bounds: 778x716
$$CH\text{1,}Q\text{19 }\left( sourse:tripos,1899 \right) \\\begin{array}{l} \,\,\text{19. If all the values of }x\,\,\text{and }y\,\,\text{given by }\\ \qquad ax^2+2hxy+by^2=\text{1,\quad }a^'x^2+2h^'xy+b^'y^2=1\\ \,\,\text{(where }a,h,b,a^',h^',b^'\,\,\text{are rational) are rational, then }\\ \qquad \left( h-h^' \right) ^2-\left( a-a^' \right) \left( b-b^' \right) ,\quad \left( ab^'-a^'b \right) ^2+4\left( ah^'-a^'h \right) \left( bh^'-b^'h \right)\\ \,\,\text{are both squares of rational numbers. }\\\end{array}\\A: \left( 1 \right) \,\, The\,\,first\,\,result\,\,is\,\,nothing\,\,but\\\,\,\left( a-a' \right) x^2+2\left( h-h' \right) xy+\left( b-b' \right) y^2=0\\\varDelta \,\,is\,\,rational\Rightarrow \left( h-h' \right) ^2-\left( a-a' \right) \left( b-b' \right) \,\,is\,\,a\,\,square\,\,of\,\,rational\,\,numbers\\\left( 2 \right) \,\,The\,\,2nd\,\,part\,\,is\,\,not\,\,that\,\,easy\\\exp\text{ress }x^2\,\,in\,\,two\,\,different\,\,ways, we\,\,have\,\,\frac{1-2hxy-by^2}{a}=\frac{1-2h'xy-b'y^2}{a'}\\\Rightarrow \left( 2ah'-2a'h \right) xy+\left( ab'-a'b \right) y^2=a-a' ①\\\text{similiraly, }\exp\text{ress }xy\,\,and\,\,y^2, we\,\,have\\\left( ah'-a'h \right) x^2+\left( bh'-b'h \right) y^2=-\left( h-h' \right) \,\, ②\\\left( ab'-a'b \right) x^2+\left( 2hb'-2bh' \right) xy=-\left( b-b' \right) \\\text{e}\lim\text{inate x , u}\sin\text{g ①②,}\\\left[ \left( ab'-a'b \right) y^2-\left( \text{a}-\text{a'} \right) \right] ^2=\left( \text{2ah'}-\text{2a'h} \right) ^2\text{x}^2\text{y}^2\\=4\left( \text{ah'}-\text{a'h} \right) ^2\text{y}^2\left[ -\frac{\left( \text{bh'}-\text{b'h} \right) \text{y}^2+\left( \text{h}-\text{h'} \right)}{\text{ah'}-\text{a'h}} \right] \\=-4\left( \text{ah'}-\text{a'h} \right) \text{y}^2\left[ \left( \text{bh'}-\text{b'h} \right) \text{y}^2+\left( \text{h}-\text{h'} \right) \right] \\\text{rearrange we have}\\\text{y}^4\left[ \left( \text{ab'}-\text{a'b} \right) ^2+4\left( \text{ah'}-\text{a'h} \right) \left( \text{bh'}-\text{b'h} \right) \right] +\text{y}^2\left[ 4\left( \text{ah'}-\text{a'h} \right) \left( \text{h}-\text{h'} \right) -\left( ab'-a'b \right) \left( \text{a}-\text{a'} \right) \right] +\left( \text{a}-\text{a'} \right) ^2=0\\\text{let m}=\left( \text{ab'}-\text{a'b} \right) ^2+4\left( \text{ah'}-\text{a'h} \right) \left( \text{bh'}-\text{b'h} \right) ,\text{n}=4\left( \text{ah'}-\text{a'h} \right) \left( \text{h}-\text{h'} \right) -\left( ab'-a'b \right) \left( \text{a}-\text{a'} \right) ,\text{t}=\left( \text{a}-\text{a'} \right) \\\text{thus my}^4+\text{ny}^2+\text{t}^2=0\\\text{then y}_1^2\text{y}_2^2=\frac{\text{t}^2}{\text{m}}\Rightarrow \text{m}=\left( \text{y}_1\text{y}_2\text{t} \right) ^2\Rightarrow \text{m is a square of rational number}\\\text{so }\left( ab^'-a^'b \right) ^2+4\left( ah^'-a^'h \right) \left( bh^'-b^'h \right) \,\,\text{is a square of rational number}$$
i think it wants you to split this into smaller bits.
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#18
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#19
Thanks!
(Original post by the bear)
i think it wants you to split this into smaller bits.
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