# Number theory

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The A.G

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x^2 + y^2 = z^2. Prove xyz is a multiple of 60

I’ve searched for the solution but almost every one uses modular arithmetic and quadratic residues. Is there a way of solving this question without knowing those things. My approach was first to consider the prime factorisation of 60=3x4x5 I also noted for the base case of a Pythagorean triplet like 3,4,5 any scalar multiple will result with xyz=60 however that doesn’t prove the question. I tried doing something with odd and even but kind of stuck

I’ve searched for the solution but almost every one uses modular arithmetic and quadratic residues. Is there a way of solving this question without knowing those things. My approach was first to consider the prime factorisation of 60=3x4x5 I also noted for the base case of a Pythagorean triplet like 3,4,5 any scalar multiple will result with xyz=60 however that doesn’t prove the question. I tried doing something with odd and even but kind of stuck

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mqb2766

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#2

(Original post by

x^2 + y^2 = z^2. Prove xyz is a multiple of 60

I’ve searched for the solution but almost every one uses modular arithmetic and quadratic residues. Is there a way of solving this question without knowing those things. My approach was first to consider the prime factorisation of 60=3x4x5 I also noted for the base case of a Pythagorean triplet like 3,4,5 any scalar multiple will result with xyz=60 however that doesn’t prove the question. I tried doing something with odd and even but kind of stuck

**The A.G**)x^2 + y^2 = z^2. Prove xyz is a multiple of 60

I’ve searched for the solution but almost every one uses modular arithmetic and quadratic residues. Is there a way of solving this question without knowing those things. My approach was first to consider the prime factorisation of 60=3x4x5 I also noted for the base case of a Pythagorean triplet like 3,4,5 any scalar multiple will result with xyz=60 however that doesn’t prove the question. I tried doing something with odd and even but kind of stuck

https://en.wikipedia.org/wiki/Pythagorean_triple

For primitives, m and n are coprime and not both odd and k=1.

Not work it through fully, but the 4 factor is easy (primitive) then its "just" about arguing a, b or c has factors of 3 and 5 using m^2 +/- n^2 and 2mn. Which is not too bad but needs a small numer of mod (remainder) arguments.

Edit

There are a couple of solutions (using relatively simple mod / FLT) in

https://math.stackexchange.com/quest...b2/53183#53183

Last edited by mqb2766; 1 month ago

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The A.G

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You could do the Euclid - pythagorean triplet representation for a,b,c in terms of m and n, then think about the product abc in terms of m and n (and k)

https://en.wikipedia.org/wiki/Pythagorean_triple

For primitives, m and n are coprime and not both odd and k=1.

Not work it through fully, but the 4 factor is easy (primitive) then its "just" about arguing a, b or c has factors of 3 and 5 using m^2 +/- n^2 and 2mn. Which is not too bad but needs a small numer of mod arguments.

Edit

There are a couple of solutions (using relatively simple mod / FLT) in

https://math.stackexchange.com/quest...b2/53183#53183

**mqb2766**)You could do the Euclid - pythagorean triplet representation for a,b,c in terms of m and n, then think about the product abc in terms of m and n (and k)

https://en.wikipedia.org/wiki/Pythagorean_triple

For primitives, m and n are coprime and not both odd and k=1.

Not work it through fully, but the 4 factor is easy (primitive) then its "just" about arguing a, b or c has factors of 3 and 5 using m^2 +/- n^2 and 2mn. Which is not too bad but needs a small numer of mod arguments.

Edit

There are a couple of solutions (using relatively simple mod / FLT) in

https://math.stackexchange.com/quest...b2/53183#53183

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mqb2766

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Would it be possible to solve it without any knowledge of mods? Is it worth learning the basics of modular arithmetic and their applications in questions for the interview?

**The A.G**)Would it be possible to solve it without any knowledge of mods? Is it worth learning the basics of modular arithmetic and their applications in questions for the interview?

n^2 = 0 or 1 mod 4

So every square number is remainder 0 or 1 on division by 4. Its an elementary odd/even proof if necessary. Id not try and learn a new technique for an interview (unless specifically required). Rather do something that you find interesting and enthuse about that.

For instance, if you're happy about Euclids pythagorean triple stuff, then to prove the 3 and 5 factor as well you can simply argue something like either 3 divides m or n, or if not then both m^4 and n^4 = 1 mod 3 so the difference will have 0 remainder on division by 3. Similarly for 5 factor. At the simplest level, you're simply talking about remainders.

An off the cuff remark made to someone recently was that remainders (modulo arithmetic) was taught in primary/junior school and university. Its not really covered very much in the years inbetween. Even though it is very important at university. A working knowledge of remainders like

n^2 = 0 or 1 mod 4

n^2 = 0 or 1 mod 3

and why this occurs would be the useful thing to know/prove.

Last edited by mqb2766; 1 month ago

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The A.G Don't know if youve come across it, but

https://books.openbookpublishers.com...7/obp.0168.pdf

is quite a good review/overview (lots of "simple" exercises) of stuff you should be aware of. There is an early section (1.4?) on pythagorean triples (Euclid) showing the 2mn, m^2+/-n^2 as well as other stuff like continued fractions, ... Also, I like Excursions in Number Theory - Ogilvy. Its old but readable, relatively short and cheap. Its less about new stuff (it has pascals triangle, continued fractions, pythagoras, sqrt(2), series telescoping, ...) but about thinking about the relationship between numbers.

If you remember one of the mat questions you had was related to

4xy = (x+y)^2 - (x-y)^2

which is also related to the am-gm inequality. So a simple difference of squares. This is also relevant here as if you considered square numbers x=m^2, y=n^2 then you'd have

(m^2 - n^2)^2 + 4m^2n^2 = (m^2+n^2)^2

and let a = m^2-n^2 and b=2mn and c^2 = a^2+b^2

a^2 + b^2 = c^2

So there is a direct relationship between a simple combination of binomials and pythagorus. Its less about learning new "number theory", but about thinking about stuff you "already" know.

Slighly repeating the previous post, for the question above using this approach, you have

n^2 = 0 or 1 mod 3

n^4 = 0 or 1 mod 5

and these two results (with the divisible by 4) give you result. You could recognise that its Fermats little theorem https://en.wikipedia.org/wiki/Fermat%27s_little_theorem or more simply just show that when you square a number, it has remainder 0 or 1 when divided by 3 and the fourth power has remainder 0 or 1 when divided by 5. This then gives you your result as the product involves (m^2-n^2) and (m^4-n^4). I doubt you'd be asked closely about FLT, but it would be fair game to think about remainders when you square, cube, ... a number.

https://books.openbookpublishers.com...7/obp.0168.pdf

is quite a good review/overview (lots of "simple" exercises) of stuff you should be aware of. There is an early section (1.4?) on pythagorean triples (Euclid) showing the 2mn, m^2+/-n^2 as well as other stuff like continued fractions, ... Also, I like Excursions in Number Theory - Ogilvy. Its old but readable, relatively short and cheap. Its less about new stuff (it has pascals triangle, continued fractions, pythagoras, sqrt(2), series telescoping, ...) but about thinking about the relationship between numbers.

If you remember one of the mat questions you had was related to

4xy = (x+y)^2 - (x-y)^2

which is also related to the am-gm inequality. So a simple difference of squares. This is also relevant here as if you considered square numbers x=m^2, y=n^2 then you'd have

(m^2 - n^2)^2 + 4m^2n^2 = (m^2+n^2)^2

and let a = m^2-n^2 and b=2mn and c^2 = a^2+b^2

a^2 + b^2 = c^2

So there is a direct relationship between a simple combination of binomials and pythagorus. Its less about learning new "number theory", but about thinking about stuff you "already" know.

Slighly repeating the previous post, for the question above using this approach, you have

n^2 = 0 or 1 mod 3

n^4 = 0 or 1 mod 5

and these two results (with the divisible by 4) give you result. You could recognise that its Fermats little theorem https://en.wikipedia.org/wiki/Fermat%27s_little_theorem or more simply just show that when you square a number, it has remainder 0 or 1 when divided by 3 and the fourth power has remainder 0 or 1 when divided by 5. This then gives you your result as the product involves (m^2-n^2) and (m^4-n^4). I doubt you'd be asked closely about FLT, but it would be fair game to think about remainders when you square, cube, ... a number.

Last edited by mqb2766; 1 month ago

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