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to prove something is a square of rational

 If all the values of x and y given by ax2+2hxy+by2=1,a′x2+2h′xy+b′y2=1 (where a,h,b,a′,h′,b′ are rational) are rational, then (h−h′)2−(a−a′)(b−b′),(ab′−a′b)2+4(ah′−a′h)(bh′−b′h) are both squares of rational numbers. \begin{array}{l}{\text { If all the values of } x \text { and } y \text { given by }} \\ {\qquad a x^{2}+2 h x y+b y^{2}=1, \quad a^{\prime} x^{2}+2 h^{\prime} x y+b^{\prime} y^{2}=1} \\ {\text { (where } a, h, b, a^{\prime}, h^{\prime}, b^{\prime} \text { are rational) are rational, then }} \\ {\qquad\left(h-h^{\prime}\right)^{2}-\left(a-a^{\prime}\right)\left(b-b^{\prime}\right), \quad\left(a b^{\prime}-a^{\prime} b\right)^{2}+4\left(a h^{\prime}-a^{\prime} h\right)\left(b h^{\prime}-b^{\prime} h\right)} \\ {\text { are both squares of rational numbers. }}\end{array}
Reply 1
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cxs
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The first part is very obvious, but I have no idea how to do the second part
help!!!
Reply 2
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max271
5
There may be a less laborious method, but have you tried writing ax2=1−2hxy−by2ax^2=1-2hxy-by^2 and b′y2=1−2h′xy−a′x2b^{\prime}y^2=1-2h^{\prime}xy-a^{\prime}x^2. Multiplying these together, we obtain ab′x2y2ab^{\prime}x^2y^2. Doing the same for a′bx2y2a^{\prime}bx^2y^2 and subtracting the two, we obtain 2(ab′−ba′)x2y2=(a−a′)x2+(b′−b)y2+2(ha′−ah′)x3y+2(h′b−b′h)y3x2(ab^{\prime}-ba^{\prime})x^2y^2=(a-a^{\prime})x^2+(b^{\prime}-b)y^2+2(ha^{\prime}-ah^{\prime})x^3y+2(h^{\prime}b-b^{\prime}h)y^3x which is beginning to look promising.
(edited 4 years ago)
Reply 3
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mqb2766
21
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
 If all the values of x and y given by ax2+2hxy+by2=1,a′x2+2h′xy+b′y2=1 (where a,h,b,a′,h′,b′ are rational) are rational, then (h−h′)2−(a−a′)(b−b′),(ab′−a′b)2+4(ah′−a′h)(bh′−b′h) are both squares of rational numbers. \begin{array}{l}{\text { If all the values of } x \text { and } y \text { given by }} \\ {\qquad a x^{2}+2 h x y+b y^{2}=1, \quad a^{\prime} x^{2}+2 h^{\prime} x y+b^{\prime} y^{2}=1} \\ {\text { (where } a, h, b, a^{\prime}, h^{\prime}, b^{\prime} \text { are rational) are rational, then }} \\ {\qquad\left(h-h^{\prime}\right)^{2}-\left(a-a^{\prime}\right)\left(b-b^{\prime}\right), \quad\left(a b^{\prime}-a^{\prime} b\right)^{2}+4\left(a h^{\prime}-a^{\prime} h\right)\left(b h^{\prime}-b^{\prime} h\right)} \\ {\text { are both squares of rational numbers. }}\end{array}
If it's any consolation, it's probably from Hardy's Pure Mathematics - reprinted from the 1899 Maths. Tripos!
(edited 4 years ago)
Original post by cxs
 If all the values of x and y given by ax2+2hxy+by2=1,a′x2+2h′xy+b′y2=1 (where a,h,b,a′,h′,b′ are rational) are rational, then (h−h′)2−(a−a′)(b−b′),(ab′−a′b)2+4(ah′−a′h)(bh′−b′h) are both squares of rational numbers. \begin{array}{l}{\text { If all the values of } x \text { and } y \text { given by }} \\ {\qquad a x^{2}+2 h x y+b y^{2}=1, \quad a^{\prime} x^{2}+2 h^{\prime} x y+b^{\prime} y^{2}=1} \\ {\text { (where } a, h, b, a^{\prime}, h^{\prime}, b^{\prime} \text { are rational) are rational, then }} \\ {\qquad\left(h-h^{\prime}\right)^{2}-\left(a-a^{\prime}\right)\left(b-b^{\prime}\right), \quad\left(a b^{\prime}-a^{\prime} b\right)^{2}+4\left(a h^{\prime}-a^{\prime} h\right)\left(b h^{\prime}-b^{\prime} h\right)} \\ {\text { are both squares of rational numbers. }}\end{array}


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 :tongue:
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.
Reply 7
Avatar for cxs
cxs
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14
Yes! The book is great, except,
it does not have answers......
Original post by ghostwalker
If it's any consolation, it's probably from Hardy's Pure Mathematics - reprinted from the 1899 Maths. Tripos!
some websites compile the answers online in a PDF which you could search for
Original post by cxs
Yes! The book is great, except,
it does not have answers......
Reply 9
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cxs
OP
14
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 ax2=1−2hxy−by2ax^2=1-2hxy-by^2 and b′y2=1−2h′xy−a′x2b^{\prime}y^2=1-2h^{\prime}xy-a^{\prime}x^2. Multiplying these together, we obtain ab′x2y2ab^{\prime}x^2y^2. Doing the same for a′bx2y2a^{\prime}bx^2y^2 and subtracting the two, we obtain 2(ab′−ba′)x2y2=(a−a′)x2+(b′−b)y2+2(ha′−ah′)x3y+2(h′b−b′h)y3x2(ab^{\prime}-ba^{\prime})x^2y^2=(a-a^{\prime})x^2+(b^{\prime}-b)y^2+2(ha^{\prime}-ah^{\prime})x^3y+2(h^{\prime}b-b^{\prime}h)y^3x 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
Unparseable latex formula:

$$\left( 2ah'-2a'h \right) xy+\left( ab'-a'b \right) y^2=a-a'\\\left( ah'-a'h \right) x^2+\left( bh'-b'h \right) y^2=-\left( h-h' \right) \\\left( ab'-a'b \right) x^2+\left( 2hb'-2bh' \right) xy=-\left( b-b' \right) $$


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.
(edited 4 years ago)
Reply 10
Avatar for cxs
cxs
OP
14
??!!!!
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
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
Unparseable latex formula:

$$\left( 2ah'-2a'h \right) xy+\left( ab'-a'b \right) y^2=a-a'\\\left( ah'-a'h \right) x^2+\left( bh'-b'h \right) y^2=-\left( h-h' \right) \\\left( ab'-a'b \right) x^2+\left( 2hb'-2bh' \right) xy=-\left( b-b' \right) $$


But I do not think I can continue by this way...
Maybe I should derive functions about x^4y^4
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.
Reply 13
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cxs
OP
14
I did not manage finding the answers to the whole book. @Satyr
But I find this!!!
https://math.stackexchange.com/questions/2537743/show-that-ab-ab2-4ah-ahbh-bh-is-a-perfect-square?r=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.
(edited 4 years ago)
Reply 14
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cxs
OP
14
It turns out to be a really hard question.
Reply 15
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cxs
OP
14
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 latex formula:

$$A: \left( 1 \right) \,\, The\,\,first\,\,result\,\,is\,\,nothing\,\,but\\\,\,\left( a-a' \right) x^2+2\left( h-h' \right) xy+\left( b-b' \right) y^2=0\\\varDelta \,\,is\,\,rational\Rightarrow \left( h-h' \right) ^2-\left( a-a' \right) \left( b-b' \right) \,\,is\,\,a\,\,square\,\,of\,\,rational\,\,numbers\\\left( 2 \right) \,\,The\,\,2nd\,\,part\,\,is\,\,not\,\,that\,\,easy\\\exp\text{ress }x^2\,\,in\,\,two\,\,different\,\,ways, we\,\,have\,\,\frac{1-2hxy-by^2}{a}=\frac{1-2h'xy-b'y^2}{a'}\\\Rightarrow \left( 2ah'-2a'h \right) xy+\left( ab'-a'b \right) y^2=a-a' ①\\\text{similiraly, }\exp\text{ress }xy\,\,and\,\,y^2, we\,\,have\\\left( ah'-a'h \right) x^2+\left( bh'-b'h \right) y^2=-\left( h-h' \right) \,\, ②\\\left( ab'-a'b \right) x^2+\left( 2hb'-2bh' \right) xy=-\left( b-b' \right) \\\text{e}\lim\text{inate x , u}\sin\text{g ①②,}$$

(edited 4 years ago)
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 latex formula:

$$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.
Reply 17
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cxs
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14
Unparseable latex formula:

\\\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}

Reply 18
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cxs
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14
Thanks!
Original post by the bear
i think it wants you to split this into smaller bits.

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