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henpen
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Is this true: 'The only way you can get from one perfect square, q^2, to another, p^2, by multiplying by a constant [i.e. p^2=a^2 q^2] is if q IS that factor [i.e. q^2=a^2]'?
 one year ago
 one year ago
henpen Group Title
Is this true: 'The only way you can get from one perfect square, q^2, to another, p^2, by multiplying by a constant [i.e. p^2=a^2 q^2] is if q IS that factor [i.e. q^2=a^2]'?
 one year ago
 one year ago

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hartnn Group TitleBest ResponseYou've already chosen the best response.1
'the first square IS that factor' what is 'that' here? factor of whose ? \(b^2=ma^2\)
 one year ago

henpen Group TitleBest ResponseYou've already chosen the best response.1
\[q^2=49p^2\]Must \[q=49\]?
 one year ago

henpen Group TitleBest ResponseYou've already chosen the best response.1
I've edited it. Is that still ambiguous?
 one year ago

hartnn Group TitleBest ResponseYou've already chosen the best response.1
then i don't think its always true.... q equal 49, is not necessary
 one year ago

henpen Group TitleBest ResponseYou've already chosen the best response.1
Of course apart from the trivial case where p=1
 one year ago

henpen Group TitleBest ResponseYou've already chosen the best response.1
Counterexample: 6^2= 4 * 2^2 Damn, it's wrong!
 one year ago

hartnn Group TitleBest ResponseYou've already chosen the best response.1
yea, its false....
 one year ago

henpen Group TitleBest ResponseYou've already chosen the best response.1
This popped up in http://www.komal.hu/verseny/feladat.cgi?a=honap&h=201210&t=mat&l=en if anyone's interested...
 one year ago
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