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**NIH****Member**- Registered: 2005-06-14
- Posts: 33

Find all integer solutions of a(a + 1) = b(b + 2).

2 + 2 = 5, for large values of 2.

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**MathsIsFun****Administrator**- Registered: 2005-01-21
- Posts: 7,535

Let's try an example. How about finding b for a=1:

a(a + 1) = b(b + 2)

1(1+1) = b(b+2)

2 = b²+2b

So, let's try see if we can find a b that solves it:

b=1: b²+2b = 1²+2 = 3

b=0: b²+2b = 0

b=-1: b²+2b = (-1)²-2 = -1

b=-2: b²+2b = (-2)²-4 = 0

b=-3: b²+2b = (-3)²-6 = 3 ... oops, went straight past it

Time for thinking cap ...

"The physicists defer only to mathematicians, and the mathematicians defer only to God ..." - Leon M. Lederman

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**kylekatarn****Member**- Registered: 2005-07-24
- Posts: 445

integer equations!

these one's are always a challenge=P

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**kylekatarn****Member**- Registered: 2005-07-24
- Posts: 445

a=0 b=0

until now I have only found the solution a=0 and b=0

probably there aren't more solutions

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**MathsIsFun****Administrator**- Registered: 2005-01-21
- Posts: 7,535

Is it possible to use an odds and evens approach ?

If a is odd then a(a+1) will be odd×even = even

If a is even then a(a+1) will be even×odd = even

If b is odd then b(b+2) will be odd×odd = odd

If b is even then b(b+2) will be even×even = even

Ooops ... gotta go ...

"The physicists defer only to mathematicians, and the mathematicians defer only to God ..." - Leon M. Lederman

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**ganesh****Moderator**- Registered: 2005-06-28
- Posts: 13,439

Thanks, MathsisFun.

I'll continue from where you left.

a(a+1) is certainly even.

b(b+2) is odd if b is odd.

If b is even, b(b+2) is even, and is always a multiple of 4.

a(a+1) is a multiple of 4 only if a is multiple of 4 or it is a number of the form 4n+3.

Lets assume a is a number of the form 4n+3.

a(a+1) = (4n+3)(4n+4) = 16n² + 16n + 12n + 12 = 16n² + 28n + 12

= 4(4n²+7n+3)

Let this number be equal to b(b+2)

b(b+2) = 4(4n²+7n+3)

b² + 2b - 4(4n²+7n+3) = 0

b = [-2 ± √{4 +16(4n²+7n+3)}]/4

Let n=0, we get an irrational number as b.

Let n=1, we get an irrational number as b.

Let n=2, again, we get an irrational number as b.

Can {4 +16(4n²+7n+3)} never be a perfect square?

Is it because the last digit of {4 +16(4n²+7n+3)} is 2 or 8 and not 4 or 6?

*Last edited by ganesh (2005-08-09 18:35:20)*

Character is who you are when no one is looking.

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**mathsyperson****Moderator**- Registered: 2005-06-22
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There are no squares that end in 2 or 8, so if 4 +16(4n²+7n+3) always ends in 2 or 8, then we have a proof.

Take away 4: 16(4n²+7n+3)=...8 or 4. Multiples of 16 move in cycles of 5, with the last digit being 6, 2, 8, 4, 0...

This means that 4n²+7n+3 has to always be of the form 5m+3 or 4. So how do we show that?

Why did the vector cross the road?

It wanted to be normal.

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**wcy****Member**- Registered: 2005-08-04
- Posts: 117

is it possible to use some general method as outlined by Euclid

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**NIH****Member**- Registered: 2005-06-14
- Posts: 33

mathsyperson wrote:

There are no squares that end in 2 or 8, so if 4 + 16(4n²+7n+3) always ends in 2 or 8, then we have a proof.

But if n = 3 or 4 (mod 5), then 4 + 16(4n²+7n+3) ends in 4.

Here's a hint: add 1 to both sides of the original equation.

2 + 2 = 5, for large values of 2.

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**wcy****Member**- Registered: 2005-08-04
- Posts: 117

then,

RHS=b²+2b+1= (b+1)²

but what abt LHS ?

*Last edited by wcy (2005-08-10 12:42:04)*

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**NIH****Member**- Registered: 2005-06-14
- Posts: 33

wcy wrote:

but what abt LHS ?

Then the LHS equals a² + a + 1.

For what values of a can this be a perfect square?

2 + 2 = 5, for large values of 2.

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**wcy****Member**- Registered: 2005-08-04
- Posts: 117

a can only be 0

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**NIH****Member**- Registered: 2005-06-14
- Posts: 33

wcy wrote:

a can only be 0

Or -1.

If you can show why those are the only two values, the problem is solved!

2 + 2 = 5, for large values of 2.

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**NIH****Member**- Registered: 2005-06-14
- Posts: 33

Here is a solution to this one.

a(a + 1) = b(b + 2) <=> a² + a + 1 = (b + 1)².

For positive a, a² < a² + a + 1 < (a + 1)², implying there is a perfect square between consecutive squares; contradiction.

For negative a, a(a + 1) = p(p - 1), where p = -a, so there are no solutions for p - 1 > 0; i.e., a < -1.

It is then easy to check for solutions with a = -1, 0.

These are: (a,b) = (-1,-2), (-1,0), (0,-2), (0,0).

2 + 2 = 5, for large values of 2.

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