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### Showing a Diophantine Equation Has No Solutions

```Date: 07/30/2008 at 05:03:01
From: Jessica
Subject: m^3=3n^2+3n+7

Do there exist positive integers m and n such that m^3 = 3n^2 + 3n + 7?

I'm not quite sure how to approach this. I've attempted to apply the
Diophantine Equation, except I haven't found anything that worked with
the ^3.

Another way I've considered was working with it is "mod"s. However,
again, I don't know if that works for this particular case.

```

```
Date: 07/30/2008 at 22:57:11
From: Doctor Ali
Subject: Re: m^3=3n^2+3n+7

Hi Jessica!

Thanks for writing to Dr. Math.  It sounds like you've done some good
thinking so far.

I'm going to prove that there exists no answer to the equation.
going on.

Let's write the equation as,

m^3 = 3 n (n + 1) + 7

Does that make sense?

On the right-hand side, we have '3 n (n+1)'.  It is surely divisible
by 3 since we see a 3 being multiplied by it.  Also we know that
surely, one of n or (n+1) is even.  So, we can deduce that the
expression is divisible by 3x2 or 6.  Do you agree?

Now, let's write 7 as (6+1).  So, we already have,

m^3   =   3 n (n + 1) + 6 + 1
\_____________/
Divisible by 6

We know that six different cases can happen to m when divided by
six.  That is,

m = 0 (mod 6)
m = 1 (mod 6)
m = 2 (mod 6)
m = 3 (mod 6)
m = 4 (mod 6)
m = 5 (mod 6)

Now, let's raise the sides to power three to see what are the
remainders of m^3 (cubes) when divided by six.

m^3 = 0^3 = 0 (mod 6)
m^3 = 1^3 = 1 (mod 6)
m^3 = 2^3 = 8 = 2 (mod 6)
m^3 = 3^3 = 27 = 3 (mod 6)
m^3 = 4^3 = 64 = 4 (mod 6)
m^3 = 5^3 = 125 = 5 (mod 6)

So, we can deduce that m has to have remainder 1 when divided by 6.
This is because we saw that the right-hand side has remainder of 1
when divided by 6.  So, if we are going to have any solutions, in all
of them we have to have

m = 1 (mod 6) .

So, we can write,

m = 6 k + 1

Let's plug it in the equation.  We'll have,

(6k + 1)^3 = 3 n (n + 1) + 7

If you expand the left-hand side and simplify the equality a little
bit, you'll have,

216 k^3 + 108 k^2 + 18 k = 3 n^2 + 3 n + 6

We can divide both sides by three and write the equation as,

6 k (12 k^2 + 6 k + 1) = n^2 + n + 2

As you see, the left-hand side is divisible by 3.  I'm going to prove
that the right-hand side is never divisible by 3 and this last step
will finish the proof.

We know that n can have three different cases when divided by 3.
That is,

n = 0 (mod 3)
n = 1 (mod 3)
n = 2 (mod 3)

Do you agree?

Now, let's build the right hand side, namely 'n^2 + n + 2' to see
what remainders are possible for 'n^2 + n + 2' when divided by 3.

for n = 0, n^2 + n + 2 = 0^2 + 0 + 2 = 2 (mod 3)
for n = 1, n^2 + n + 2 = 1^2 + 1 + 2 = 1 (mod 3)
for n = 2, n^2 + n + 2 = 2^2 + 2 + 2 = 2 (mod 3)

As you see, none of the above lead to zero remainder when divided by
three.  This means that the right-hand side is never divisible by
three.  This is while we saw that the left-hand side is divisible by

Does that make sense?  Please write back if you still have any
difficulties.

- Doctor Ali, The Math Forum
http://mathforum.org/dr.math/
```
Associated Topics:
High School Number Theory

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