# to prove something is a square of rational

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The first part is very obvious, but I have no idea how to do the second part

help!!!

help!!!

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#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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#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.

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.

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#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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#6

(Original post by

If it's an consolation, it's probably from Hardy's Pure Mathematics - reprinted from the 1899 Maths. Tripos!

**ghostwalker**)If it's an consolation, it's probably from Hardy's Pure Mathematics - reprinted from the 1899 Maths. Tripos!

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#7

(Original post by

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.

**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.

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Yes! The book is great, except,

it does not have answers......

it does not have answers......

(Original post by

If it's any consolation, it's probably from Hardy's Pure Mathematics - reprinted from the 1899 Maths. Tripos!

**ghostwalker**)If it's any consolation, it's probably from Hardy's Pure Mathematics - reprinted from the 1899 Maths. Tripos!

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I tried this but, how to go further???

Feel like I lost some key points.

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

Feel like I lost some key points.

(Original post by

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.

**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.

But I do not think I can continue in this way...

Maybe I should derive functions about x^4y^4

(Original post by

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.

**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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??!!!!

Perhaps I just did a rough search weeks ago... I should search again

Perhaps I just did a rough search weeks ago... I should search again

(Original post by

some websites compile the answers online in a PDF which you could search for

**Satyr**)some websites compile the answers online in a PDF which you could search for

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#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

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

**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

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#13

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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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.

But I find this!!!

https://math.stackexchange.com/quest...=SearchResults

It is solved in a really brilliant way.

(Original post by

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.

**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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#17

(Original post by

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.

**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}$$`

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Thanks!

(Original post by

i think it wants you to split this into smaller bits.

**the bear**)i think it wants you to split this into smaller bits.

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