Solving equations depending on $z$

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For witch real number $z$, are there positive integers $a$, $b$ and $c$, that fulfill this system of equations? $(a,b,c)∈ℕ$

$a+b+c=57$

$a^2+b^2−c^2=z$

$z·c=2017$

I need to determine all variables $(a, b, c)$ depending on $z$.
Any hints?

edited Nov 23 at 12:01

Harry Peter

5,43111439

asked Sep 13 at 14:05

calculatormathematical

389

1

Note that $z$ must also be an integer if $a,b,c$ are. Now if $c cdot z = 2017$ and $2017$ is prime,..
– астон вілла олоф мэллбэрг
Sep 13 at 14:08

$(a+b)^2 neq a^2 + b^2,$ similarly for $(57+z)^2.$
– user376343
Sep 13 at 14:23

"Now one can eliminate $c^2$..." Sure?
– Paul
Sep 13 at 14:30

But $a^2+b^2 = 57^2-c^2$
– calculatormathematical
Sep 13 at 14:57

add a comment |

up vote
0
down vote

favorite

For witch real number $z$, are there positive integers $a$, $b$ and $c$, that fulfill this system of equations? $(a,b,c)∈ℕ$

$a+b+c=57$

$a^2+b^2−c^2=z$

$z·c=2017$

I need to determine all variables $(a, b, c)$ depending on $z$.
Any hints?

edited Nov 23 at 12:01

Harry Peter

5,43111439

asked Sep 13 at 14:05

calculatormathematical

389

1

Note that $z$ must also be an integer if $a,b,c$ are. Now if $c cdot z = 2017$ and $2017$ is prime,..
– астон вілла олоф мэллбэрг
Sep 13 at 14:08

$(a+b)^2 neq a^2 + b^2,$ similarly for $(57+z)^2.$
– user376343
Sep 13 at 14:23

"Now one can eliminate $c^2$..." Sure?
– Paul
Sep 13 at 14:30

But $a^2+b^2 = 57^2-c^2$
– calculatormathematical
Sep 13 at 14:57

add a comment |

up vote
0
down vote

favorite

For witch real number $z$, are there positive integers $a$, $b$ and $c$, that fulfill this system of equations? $(a,b,c)∈ℕ$

$a+b+c=57$

$a^2+b^2−c^2=z$

$z·c=2017$

I need to determine all variables $(a, b, c)$ depending on $z$.
Any hints?

edited Nov 23 at 12:01

Harry Peter

5,43111439

asked Sep 13 at 14:05

calculatormathematical

389

For witch real number $z$, are there positive integers $a$, $b$ and $c$, that fulfill this system of equations? $(a,b,c)∈ℕ$

$a+b+c=57$

$a^2+b^2−c^2=z$

$z·c=2017$

I need to determine all variables $(a, b, c)$ depending on $z$.
Any hints?

systems-of-equations diophantine-equations quadratics

edited Nov 23 at 12:01

Harry Peter

5,43111439

asked Sep 13 at 14:05

calculatormathematical

389

edited Nov 23 at 12:01

Harry Peter

5,43111439

asked Sep 13 at 14:05

calculatormathematical

389

edited Nov 23 at 12:01

Harry Peter

5,43111439

edited Nov 23 at 12:01

Harry Peter

5,43111439

edited Nov 23 at 12:01

Harry Peter

5,43111439

asked Sep 13 at 14:05

calculatormathematical

389

asked Sep 13 at 14:05

calculatormathematical

389

asked Sep 13 at 14:05

calculatormathematical

389

1

Note that $z$ must also be an integer if $a,b,c$ are. Now if $c cdot z = 2017$ and $2017$ is prime,..
– астон вілла олоф мэллбэрг
Sep 13 at 14:08

$(a+b)^2 neq a^2 + b^2,$ similarly for $(57+z)^2.$
– user376343
Sep 13 at 14:23

"Now one can eliminate $c^2$..." Sure?
– Paul
Sep 13 at 14:30

But $a^2+b^2 = 57^2-c^2$
– calculatormathematical
Sep 13 at 14:57

add a comment |

1

Note that $z$ must also be an integer if $a,b,c$ are. Now if $c cdot z = 2017$ and $2017$ is prime,..
– астон вілла олоф мэллбэрг
Sep 13 at 14:08

$(a+b)^2 neq a^2 + b^2,$ similarly for $(57+z)^2.$
– user376343
Sep 13 at 14:23

"Now one can eliminate $c^2$..." Sure?
– Paul
Sep 13 at 14:30

But $a^2+b^2 = 57^2-c^2$
– calculatormathematical
Sep 13 at 14:57

Note that $z$ must also be an integer if $a,b,c$ are. Now if $c cdot z = 2017$ and $2017$ is prime,..
– астон вілла олоф мэллбэрг
Sep 13 at 14:08

$(a+b)^2 neq a^2 + b^2,$ similarly for $(57+z)^2.$
– user376343
Sep 13 at 14:23

"Now one can eliminate $c^2$..." Sure?
– Paul
Sep 13 at 14:30

But $a^2+b^2 = 57^2-c^2$
– calculatormathematical
Sep 13 at 14:57

add a comment |

1 Answer
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0
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I have found a solution for my problem.

There are no numbers that fulfill the equation.
So (a,b,c) ∈ ℕ

$z*c=2017$

2017 is a Prime number. If

$z ≠ 1 ⋁ 2017$

then $c$ will logically be an odd number,
because else if z is not 1 or 2017 we'll have:

$2017/z = c$ (≠ even)

So $z$ has to be definitely 1 or 2017.

So we have case one: c = 1 if z is 2017
and case two: c = 2017 if z is 1.

Case one:
$(c=1;z=2017)$

1.) $a+b-1 = 57$
2.) $a^2 +b^2 -c^2= 2017$

If we divide 2.) by 1.)
we get:

1.) $a + b + 1 = 57 => a+b = 56$

$a^2 + b^2 -1^2 = 2017 => a^2 + b^2 = 2018$

2.) / 1.) $(a^2 +b^2) / (a+b) = 2018 / 56$

This makes no sense... ↯

Case two:
$(c=2017;z=1)$

1.) $a+b +2017 = 57$

2.) $a^2 + b^2 - 2017^2 = 1$

We end up with:

$(a^2 + b^2) / (a+b) = (2017^2) / (57-2017)$

I'm not 100% sure but i guess i'm right.
If not, pls comment.

Thx.

answered Nov 13 at 20:46

community wiki

calculatormathematical

add a comment |

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1 Answer
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1 Answer
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oldest

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up vote
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I have found a solution for my problem.

There are no numbers that fulfill the equation.
So (a,b,c) ∈ ℕ

$z*c=2017$

2017 is a Prime number. If

$z ≠ 1 ⋁ 2017$

then $c$ will logically be an odd number,
because else if z is not 1 or 2017 we'll have:

$2017/z = c$ (≠ even)

So $z$ has to be definitely 1 or 2017.

So we have case one: c = 1 if z is 2017
and case two: c = 2017 if z is 1.

Case one:
$(c=1;z=2017)$

1.) $a+b-1 = 57$
2.) $a^2 +b^2 -c^2= 2017$

If we divide 2.) by 1.)
we get:

1.) $a + b + 1 = 57 => a+b = 56$

$a^2 + b^2 -1^2 = 2017 => a^2 + b^2 = 2018$

2.) / 1.) $(a^2 +b^2) / (a+b) = 2018 / 56$

This makes no sense... ↯

Case two:
$(c=2017;z=1)$

1.) $a+b +2017 = 57$

2.) $a^2 + b^2 - 2017^2 = 1$

We end up with:

$(a^2 + b^2) / (a+b) = (2017^2) / (57-2017)$

I'm not 100% sure but i guess i'm right.
If not, pls comment.

Thx.

answered Nov 13 at 20:46

community wiki

calculatormathematical

add a comment |

up vote
0
down vote

I have found a solution for my problem.

There are no numbers that fulfill the equation.
So (a,b,c) ∈ ℕ

$z*c=2017$

2017 is a Prime number. If

$z ≠ 1 ⋁ 2017$

then $c$ will logically be an odd number,
because else if z is not 1 or 2017 we'll have:

$2017/z = c$ (≠ even)

So $z$ has to be definitely 1 or 2017.

So we have case one: c = 1 if z is 2017
and case two: c = 2017 if z is 1.

Case one:
$(c=1;z=2017)$

1.) $a+b-1 = 57$
2.) $a^2 +b^2 -c^2= 2017$

If we divide 2.) by 1.)
we get:

1.) $a + b + 1 = 57 => a+b = 56$

$a^2 + b^2 -1^2 = 2017 => a^2 + b^2 = 2018$

2.) / 1.) $(a^2 +b^2) / (a+b) = 2018 / 56$

This makes no sense... ↯

Case two:
$(c=2017;z=1)$

1.) $a+b +2017 = 57$

2.) $a^2 + b^2 - 2017^2 = 1$

We end up with:

$(a^2 + b^2) / (a+b) = (2017^2) / (57-2017)$

I'm not 100% sure but i guess i'm right.
If not, pls comment.

Thx.

answered Nov 13 at 20:46

community wiki

calculatormathematical

add a comment |

up vote
0
down vote

I have found a solution for my problem.

There are no numbers that fulfill the equation.
So (a,b,c) ∈ ℕ

$z*c=2017$

2017 is a Prime number. If

$z ≠ 1 ⋁ 2017$

then $c$ will logically be an odd number,
because else if z is not 1 or 2017 we'll have:

$2017/z = c$ (≠ even)

So $z$ has to be definitely 1 or 2017.

So we have case one: c = 1 if z is 2017
and case two: c = 2017 if z is 1.

Case one:
$(c=1;z=2017)$

1.) $a+b-1 = 57$
2.) $a^2 +b^2 -c^2= 2017$

If we divide 2.) by 1.)
we get:

1.) $a + b + 1 = 57 => a+b = 56$

$a^2 + b^2 -1^2 = 2017 => a^2 + b^2 = 2018$

2.) / 1.) $(a^2 +b^2) / (a+b) = 2018 / 56$

This makes no sense... ↯

Case two:
$(c=2017;z=1)$

1.) $a+b +2017 = 57$

2.) $a^2 + b^2 - 2017^2 = 1$

We end up with:

$(a^2 + b^2) / (a+b) = (2017^2) / (57-2017)$

I'm not 100% sure but i guess i'm right.
If not, pls comment.

Thx.

answered Nov 13 at 20:46

community wiki

calculatormathematical

I have found a solution for my problem.

There are no numbers that fulfill the equation.
So (a,b,c) ∈ ℕ

$z*c=2017$

2017 is a Prime number. If

$z ≠ 1 ⋁ 2017$

then $c$ will logically be an odd number,
because else if z is not 1 or 2017 we'll have:

$2017/z = c$ (≠ even)

So $z$ has to be definitely 1 or 2017.

So we have case one: c = 1 if z is 2017
and case two: c = 2017 if z is 1.

Case one:
$(c=1;z=2017)$

1.) $a+b-1 = 57$
2.) $a^2 +b^2 -c^2= 2017$

If we divide 2.) by 1.)
we get:

1.) $a + b + 1 = 57 => a+b = 56$

$a^2 + b^2 -1^2 = 2017 => a^2 + b^2 = 2018$

2.) / 1.) $(a^2 +b^2) / (a+b) = 2018 / 56$

This makes no sense... ↯

Case two:
$(c=2017;z=1)$

1.) $a+b +2017 = 57$

2.) $a^2 + b^2 - 2017^2 = 1$

We end up with:

$(a^2 + b^2) / (a+b) = (2017^2) / (57-2017)$

I'm not 100% sure but i guess i'm right.
If not, pls comment.

Thx.

answered Nov 13 at 20:46

community wiki

calculatormathematical

answered Nov 13 at 20:46

community wiki

calculatormathematical

community wiki

calculatormathematical

community wiki

calculatormathematical

add a comment |

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