Mixing
Linking Algebra and Number Theory
$begingroup$
a,b,c is an real number. Now, let's say that p(x)=ax^2+bx+c. Then,
find an equivalent condition that always has an integer value(the p(x)'s value) for an arbitrary integer x.
Sorry guys I don't know MathJax.
Anyway, this problem is represented in Hungary Eotvos 1902.
number-theory contest-math
asked Jan 19 at 8:40
mathheromathhero
235
$endgroup$
add a comment |
$begingroup$
a,b,c is an real number. Now, let's say that p(x)=ax^2+bx+c. Then,
find an equivalent condition that always has an integer value(the p(x)'s value) for an arbitrary integer x.
Sorry guys I don't know MathJax.
Anyway, this problem is represented in Hungary Eotvos 1902.
number-theory contest-math
asked Jan 19 at 8:40
mathheromathhero
235
$endgroup$
add a comment |
$begingroup$
a,b,c is an real number. Now, let's say that p(x)=ax^2+bx+c. Then,
find an equivalent condition that always has an integer value(the p(x)'s value) for an arbitrary integer x.
Sorry guys I don't know MathJax.
Anyway, this problem is represented in Hungary Eotvos 1902.
number-theory contest-math
asked Jan 19 at 8:40
mathheromathhero
235
$endgroup$
a,b,c is an real number. Now, let's say that p(x)=ax^2+bx+c. Then,
find an equivalent condition that always has an integer value(the p(x)'s value) for an arbitrary integer x.
Sorry guys I don't know MathJax.
Anyway, this problem is represented in Hungary Eotvos 1902.
number-theory contest-math
number-theory contest-math
asked Jan 19 at 8:40
mathheromathhero
235
asked Jan 19 at 8:40
mathheromathhero
235
asked Jan 19 at 8:40
mathheromathhero
235
asked Jan 19 at 8:40
mathheromathhero
235
asked Jan 19 at 8:40
mathheromathhero
235
235
add a comment |
add a comment |
1 Answer
1
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$begingroup$
If you put $x = -1, 0, 1$, then you get $cin mathbb{Z}$, $a+b, a-bin mathbb{Z}$, which implies $a = frac{m}{2}, b = frac{n}{2}$ for some $m, nin mathbb{Z}$ with $mequiv n(mathrm{mod},2)$. Now try to show the converse - if $a, b, c$ are of this form, then $f(x)in mathbb{Z}$ for any $xin mathbb{Z}$.
answered Jan 19 at 8:46
Seewoo LeeSeewoo Lee
6,959927
$endgroup$
add a comment |
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1 Answer
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1 Answer
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$begingroup$
If you put $x = -1, 0, 1$, then you get $cin mathbb{Z}$, $a+b, a-bin mathbb{Z}$, which implies $a = frac{m}{2}, b = frac{n}{2}$ for some $m, nin mathbb{Z}$ with $mequiv n(mathrm{mod},2)$. Now try to show the converse - if $a, b, c$ are of this form, then $f(x)in mathbb{Z}$ for any $xin mathbb{Z}$.
answered Jan 19 at 8:46
Seewoo LeeSeewoo Lee
6,959927
$endgroup$
add a comment |
$begingroup$
If you put $x = -1, 0, 1$, then you get $cin mathbb{Z}$, $a+b, a-bin mathbb{Z}$, which implies $a = frac{m}{2}, b = frac{n}{2}$ for some $m, nin mathbb{Z}$ with $mequiv n(mathrm{mod},2)$. Now try to show the converse - if $a, b, c$ are of this form, then $f(x)in mathbb{Z}$ for any $xin mathbb{Z}$.
answered Jan 19 at 8:46
Seewoo LeeSeewoo Lee
6,959927
$endgroup$
add a comment |
$begingroup$
If you put $x = -1, 0, 1$, then you get $cin mathbb{Z}$, $a+b, a-bin mathbb{Z}$, which implies $a = frac{m}{2}, b = frac{n}{2}$ for some $m, nin mathbb{Z}$ with $mequiv n(mathrm{mod},2)$. Now try to show the converse - if $a, b, c$ are of this form, then $f(x)in mathbb{Z}$ for any $xin mathbb{Z}$.
answered Jan 19 at 8:46
Seewoo LeeSeewoo Lee
6,959927
$endgroup$
If you put $x = -1, 0, 1$, then you get $cin mathbb{Z}$, $a+b, a-bin mathbb{Z}$, which implies $a = frac{m}{2}, b = frac{n}{2}$ for some $m, nin mathbb{Z}$ with $mequiv n(mathrm{mod},2)$. Now try to show the converse - if $a, b, c$ are of this form, then $f(x)in mathbb{Z}$ for any $xin mathbb{Z}$.
answered Jan 19 at 8:46
Seewoo LeeSeewoo Lee
6,959927
answered Jan 19 at 8:46
Seewoo LeeSeewoo Lee
6,959927
answered Jan 19 at 8:46
Seewoo LeeSeewoo Lee
6,959927
answered Jan 19 at 8:46
Seewoo LeeSeewoo Lee
6,959927
6,959927
add a comment |
add a comment |
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