Mixing
Determine all $a$ and $b$ natural numbers such that $frac {a^2+2b} {b^2-2a}$ and $frac {b^2+2a} {a^2-2b}$ are...
$begingroup$
I proceeded in the following way:
It is clear that $a ne 0$ and $b ne 0$.
Let $frac {a^2+2b} {b^2-2a} = k, k in mathbb{Z} tag 1$
and $frac {b^2+2a} {a^2-2b} = m, m in mathbb{Z} tag 2$
1. If $a = b$:
Let $a = b = n$. Then $k = m = frac {n^2+2n} {n^2-2n} = frac {n(n+2)} {n(n-2)} = frac {n+2} {n-2} = p, p in mathbb{Z}$. From here we get that $n in {1, 3, 4, 6}$. I don't know if these are all so I can't prove that there are no others.
2. If $a ne b$:
For $k$ and $m$ to be whole numbers the numerators have to be greater than or equal to the denominator in $(1)$ and $(2)$.
From $(1)$ we get that $a^2+2b ge b^2-2a implies a+2 ge b tag {3}$
From $(2)$ we get that $b^2+2a ge a^2-2b implies b+2 ge a tag {4}$
Here I am stuck again... WolframAlpha gives me these solutions to $(3)$ and $(4)$.
fractions natural-numbers
asked Jan 15 at 19:18
Krisztián KissKrisztián Kiss
484
$endgroup$
add a comment |
$begingroup$
I proceeded in the following way:
It is clear that $a ne 0$ and $b ne 0$.
Let $frac {a^2+2b} {b^2-2a} = k, k in mathbb{Z} tag 1$
and $frac {b^2+2a} {a^2-2b} = m, m in mathbb{Z} tag 2$
1. If $a = b$:
Let $a = b = n$. Then $k = m = frac {n^2+2n} {n^2-2n} = frac {n(n+2)} {n(n-2)} = frac {n+2} {n-2} = p, p in mathbb{Z}$. From here we get that $n in {1, 3, 4, 6}$. I don't know if these are all so I can't prove that there are no others.
2. If $a ne b$:
For $k$ and $m$ to be whole numbers the numerators have to be greater than or equal to the denominator in $(1)$ and $(2)$.
From $(1)$ we get that $a^2+2b ge b^2-2a implies a+2 ge b tag {3}$
From $(2)$ we get that $b^2+2a ge a^2-2b implies b+2 ge a tag {4}$
Here I am stuck again... WolframAlpha gives me these solutions to $(3)$ and $(4)$.
fractions natural-numbers
asked Jan 15 at 19:18
Krisztián KissKrisztián Kiss
484
$endgroup$
add a comment |
$begingroup$
I proceeded in the following way:
It is clear that $a ne 0$ and $b ne 0$.
Let $frac {a^2+2b} {b^2-2a} = k, k in mathbb{Z} tag 1$
and $frac {b^2+2a} {a^2-2b} = m, m in mathbb{Z} tag 2$
1. If $a = b$:
Let $a = b = n$. Then $k = m = frac {n^2+2n} {n^2-2n} = frac {n(n+2)} {n(n-2)} = frac {n+2} {n-2} = p, p in mathbb{Z}$. From here we get that $n in {1, 3, 4, 6}$. I don't know if these are all so I can't prove that there are no others.
2. If $a ne b$:
For $k$ and $m$ to be whole numbers the numerators have to be greater than or equal to the denominator in $(1)$ and $(2)$.
From $(1)$ we get that $a^2+2b ge b^2-2a implies a+2 ge b tag {3}$
From $(2)$ we get that $b^2+2a ge a^2-2b implies b+2 ge a tag {4}$
Here I am stuck again... WolframAlpha gives me these solutions to $(3)$ and $(4)$.
fractions natural-numbers
asked Jan 15 at 19:18
Krisztián KissKrisztián Kiss
484
$endgroup$
I proceeded in the following way:
It is clear that $a ne 0$ and $b ne 0$.
Let $frac {a^2+2b} {b^2-2a} = k, k in mathbb{Z} tag 1$
and $frac {b^2+2a} {a^2-2b} = m, m in mathbb{Z} tag 2$
1. If $a = b$:
Let $a = b = n$. Then $k = m = frac {n^2+2n} {n^2-2n} = frac {n(n+2)} {n(n-2)} = frac {n+2} {n-2} = p, p in mathbb{Z}$. From here we get that $n in {1, 3, 4, 6}$. I don't know if these are all so I can't prove that there are no others.
2. If $a ne b$:
For $k$ and $m$ to be whole numbers the numerators have to be greater than or equal to the denominator in $(1)$ and $(2)$.
From $(1)$ we get that $a^2+2b ge b^2-2a implies a+2 ge b tag {3}$
From $(2)$ we get that $b^2+2a ge a^2-2b implies b+2 ge a tag {4}$
Here I am stuck again... WolframAlpha gives me these solutions to $(3)$ and $(4)$.
fractions natural-numbers
fractions natural-numbers
asked Jan 15 at 19:18
Krisztián KissKrisztián Kiss
484
asked Jan 15 at 19:18
Krisztián KissKrisztián Kiss
484
edited Jan 15 at 19:26
Krisztián Kiss
asked Jan 15 at 19:18
Krisztián KissKrisztián Kiss
484
asked Jan 15 at 19:18
Krisztián KissKrisztián Kiss
484
asked Jan 15 at 19:18
Krisztián KissKrisztián Kiss
484
484
add a comment |
add a comment |
1 Answer
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$begingroup$
So we have $bin {a-2,a-1,a,a+1,a+2}$
Case 1: $b= a-2$, then $$(a-2)^2-2amid a^2+2(a-2)$$
so $$a^2-6a+4mid a^2+2a-4$$
and thus $$a^2-6a+4mid (a^2+2a-4)-(a^2-6a+4)=8a-8$$ Now if $a=1$ then $b=-1$ which works. So $a>1$ and now we have $$a^2-6a+4leq 8a-8implies a^2-14a+12leq 0$$
so $$ (a-7)^2leq 37implies |a-7|leq 6...$$
We do similary for all other cases...
answered Jan 15 at 19:37
greedoidgreedoid
43.2k1153105
$endgroup$
add a comment |
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1 Answer
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$begingroup$
So we have $bin {a-2,a-1,a,a+1,a+2}$
Case 1: $b= a-2$, then $$(a-2)^2-2amid a^2+2(a-2)$$
so $$a^2-6a+4mid a^2+2a-4$$
and thus $$a^2-6a+4mid (a^2+2a-4)-(a^2-6a+4)=8a-8$$ Now if $a=1$ then $b=-1$ which works. So $a>1$ and now we have $$a^2-6a+4leq 8a-8implies a^2-14a+12leq 0$$
so $$ (a-7)^2leq 37implies |a-7|leq 6...$$
We do similary for all other cases...
answered Jan 15 at 19:37
greedoidgreedoid
43.2k1153105
$endgroup$
add a comment |
$begingroup$
So we have $bin {a-2,a-1,a,a+1,a+2}$
Case 1: $b= a-2$, then $$(a-2)^2-2amid a^2+2(a-2)$$
so $$a^2-6a+4mid a^2+2a-4$$
and thus $$a^2-6a+4mid (a^2+2a-4)-(a^2-6a+4)=8a-8$$ Now if $a=1$ then $b=-1$ which works. So $a>1$ and now we have $$a^2-6a+4leq 8a-8implies a^2-14a+12leq 0$$
so $$ (a-7)^2leq 37implies |a-7|leq 6...$$
We do similary for all other cases...
answered Jan 15 at 19:37
greedoidgreedoid
43.2k1153105
$endgroup$
add a comment |
$begingroup$
So we have $bin {a-2,a-1,a,a+1,a+2}$
Case 1: $b= a-2$, then $$(a-2)^2-2amid a^2+2(a-2)$$
so $$a^2-6a+4mid a^2+2a-4$$
and thus $$a^2-6a+4mid (a^2+2a-4)-(a^2-6a+4)=8a-8$$ Now if $a=1$ then $b=-1$ which works. So $a>1$ and now we have $$a^2-6a+4leq 8a-8implies a^2-14a+12leq 0$$
so $$ (a-7)^2leq 37implies |a-7|leq 6...$$
We do similary for all other cases...
answered Jan 15 at 19:37
greedoidgreedoid
43.2k1153105
$endgroup$
So we have $bin {a-2,a-1,a,a+1,a+2}$
Case 1: $b= a-2$, then $$(a-2)^2-2amid a^2+2(a-2)$$
so $$a^2-6a+4mid a^2+2a-4$$
and thus $$a^2-6a+4mid (a^2+2a-4)-(a^2-6a+4)=8a-8$$ Now if $a=1$ then $b=-1$ which works. So $a>1$ and now we have $$a^2-6a+4leq 8a-8implies a^2-14a+12leq 0$$
so $$ (a-7)^2leq 37implies |a-7|leq 6...$$
We do similary for all other cases...
answered Jan 15 at 19:37
greedoidgreedoid
43.2k1153105
answered Jan 15 at 19:37
greedoidgreedoid
43.2k1153105
answered Jan 15 at 19:37
greedoidgreedoid
43.2k1153105
answered Jan 15 at 19:37
greedoidgreedoid
43.2k1153105
43.2k1153105
add a comment |
add a comment |
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