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phanthanhtom

Member

Registered: 2012-06-22

Posts: 290

Triples of integers (arithmetic)

Find all triples of integers (m, n, p)  such that m + n + p = 2002 and m^2 + n^2 + p^2 = mnp + 4.

Please provide detailed solution (with mathematical proof).

Agnishom

Real Member

From: Riemann Sphere

Registered: 2011-01-29

Posts: 24,996

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Re: Triples of integers (arithmetic)

Where is this problem from?

with mathematical proof

You are going to have trouble defining what a mathematical proof is.

Last edited by Agnishom (2015-07-19 03:51:08)

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'And fun? If maths is fun, then getting a tooth extraction is fun. A viral infection is fun. Rabies shots are fun.'
'God exists because Mathematics is consistent, and the devil exists because we cannot prove it'
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phanthanhtom

Member

Registered: 2012-06-22

Posts: 290

Re: Triples of integers (arithmetic)

Whatever.

It's from a math book without solutions. Sad but true.

bobbym

bumpkin

From: Bumpkinland

Registered: 2009-04-12

Posts: 109,606

Re: Triples of integers (arithmetic)

What is the name of the book?

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In mathematics, you don't understand things. You just get used to them.
If it ain't broke, fix it until it is.
Always satisfy the Prime Directive of getting the right answer above all else.

phanthanhtom

Member

Registered: 2012-06-22

Posts: 290

Re: Triples of integers (arithmetic)

Something along the lines of Algebraic Arithmetic (it isn't always clear how to translate into English).

Agnishom

Real Member

From: Riemann Sphere

Registered: 2011-01-29

Posts: 24,996

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Re: Triples of integers (arithmetic)

This is an interesting question, nevertheless.

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'And fun? If maths is fun, then getting a tooth extraction is fun. A viral infection is fun. Rabies shots are fun.'
'God exists because Mathematics is consistent, and the devil exists because we cannot prove it'
I'm not crazy, my mother had me tested.

Bob

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Registered: 2010-06-20

Posts: 10,621

Re: Triples of integers (arithmetic)

Anyone got one or more solutions yet?

Bob

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Children are not defined by school ...........The Fonz
You cannot teach a man anything;  you can only help him find it within himself..........Galileo Galilei
Sometimes I deliberately make mistakes, just to test you!  …………….Bob

bobbym

bumpkin

From: Bumpkinland

Registered: 2009-04-12

Posts: 109,606

Re: Triples of integers (arithmetic)

There are some known.

---

In mathematics, you don't understand things. You just get used to them.
If it ain't broke, fix it until it is.
Always satisfy the Prime Directive of getting the right answer above all else.

phanthanhtom

Member

Registered: 2012-06-22

Posts: 290

Re: Triples of integers (arithmetic)

What do you mean bobbym? You found some solutions?

bobbym

bumpkin

From: Bumpkinland

Registered: 2009-04-12

Posts: 109,606

Re: Triples of integers (arithmetic)

There are at least 3 solutions. This is not difficult to get.

---

In mathematics, you don't understand things. You just get used to them.
If it ain't broke, fix it until it is.
Always satisfy the Prime Directive of getting the right answer above all else.

phanthanhtom

Member

Registered: 2012-06-22

Posts: 290

Re: Triples of integers (arithmetic)

I have found the proof that there are only 3 solutions: (1000, 1000, 2) and its permutations.

It is very long, so I'll post it a little while later, but the key idea is: m+2, n+2 and p+2 must all be factors of 2004^2. First note that none of the three numbers m, n and p can be -1, 0 or 1 (this can be proven with discriminant). Let a=m+2, b=n+2 and c=p+2, then they are all factors of 2004^2. Consider three cases for a, b and c: all three number are positive, two are positive and one is negative, or one is positive and two are negative. We can prove that the only possible cases for a, b and c such that a+b+c=2008 are (1002,1002,4) and its permutations.

phanthanhtom

Member

Registered: 2012-06-22

Posts: 290

Re: Triples of integers (arithmetic)

I found a better solution!

2004^2 = 2002^2 + 4*2002 + 4 = (m + n + p)^2 + (m + n + p) +4 = (m^2 + n^2 + p^2) + 2(mn + np + pm) + (m+ n + p) + 4 = mnp + 4 + mn +np + pm + m + n + p + 4 = (m+2)(n+2)(p+2).