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For which integer $n$ is $28 + 101 + 2^n$ a perfect square?
$begingroup$
This question
For which integer $n$ is $$28 + 101 + 2^n$$ a perfect square. Please also suggest an algorithm to solve similar problems. Thanks
Btw, this question has been taken from an Aryabhatta exam for 8th graders in India.
elementary-number-theory square-numbers
edited Jan 31 at 15:50
Maria Mazur
50k1361124
asked Jan 31 at 15:02
Vasu090Vasu090
244
$endgroup$
add a comment |
$begingroup$
This question
For which integer $n$ is $$28 + 101 + 2^n$$ a perfect square. Please also suggest an algorithm to solve similar problems. Thanks
Btw, this question has been taken from an Aryabhatta exam for 8th graders in India.
elementary-number-theory square-numbers
edited Jan 31 at 15:50
Maria Mazur
50k1361124
asked Jan 31 at 15:02
Vasu090Vasu090
244
$endgroup$
1
$begingroup$
Try $$n=12$$ to get a square number
$endgroup$
– Dr. Sonnhard Graubner
Jan 31 at 15:09
add a comment |
$begingroup$
This question
For which integer $n$ is $$28 + 101 + 2^n$$ a perfect square. Please also suggest an algorithm to solve similar problems. Thanks
Btw, this question has been taken from an Aryabhatta exam for 8th graders in India.
elementary-number-theory square-numbers
edited Jan 31 at 15:50
Maria Mazur
50k1361124
asked Jan 31 at 15:02
Vasu090Vasu090
244
$endgroup$
This question
For which integer $n$ is $$28 + 101 + 2^n$$ a perfect square. Please also suggest an algorithm to solve similar problems. Thanks
Btw, this question has been taken from an Aryabhatta exam for 8th graders in India.
elementary-number-theory square-numbers
elementary-number-theory square-numbers
edited Jan 31 at 15:50
Maria Mazur
50k1361124
asked Jan 31 at 15:02
Vasu090Vasu090
244
edited Jan 31 at 15:50
Maria Mazur
50k1361124
asked Jan 31 at 15:02
Vasu090Vasu090
244
edited Jan 31 at 15:50
Maria Mazur
50k1361124
edited Jan 31 at 15:50
Maria Mazur
50k1361124
edited Jan 31 at 15:50
Maria Mazur
50k1361124
50k1361124
asked Jan 31 at 15:02
Vasu090Vasu090
244
asked Jan 31 at 15:02
Vasu090Vasu090
244
asked Jan 31 at 15:02
Vasu090Vasu090
244
244
1
$begingroup$
Try $$n=12$$ to get a square number
$endgroup$
– Dr. Sonnhard Graubner
Jan 31 at 15:09
add a comment |
1
$begingroup$
Try $$n=12$$ to get a square number
$endgroup$
– Dr. Sonnhard Graubner
Jan 31 at 15:09
1
1
$begingroup$
Try $$n=12$$ to get a square number
$endgroup$
– Dr. Sonnhard Graubner
Jan 31 at 15:09
$begingroup$
Try $$n=12$$ to get a square number
$endgroup$
– Dr. Sonnhard Graubner
Jan 31 at 15:09
add a comment |
4 Answers
4
active
oldest
votes
$begingroup$
You can move terms around to try to get things to look like a perfect square trinomial:
$(a + b)^2 = a^2 + 2ab + b^2$
Specifically, the special case where $a$ or $b$ is 1:
$(1 + b)^2 = 1 + 2 b + b^2$
Start by combining the constant terms, and notice that 128 is a power of 2:
$28 + 101 + 2^n = 1 + 128 + 2^n = 1 + 2 cdot 2^6 + 2^n$
We can see that $(1 + 2^6)^2 = 1 + 2 cdot 2^6 + (2^6)^2 = 1 + 2 cdot 2^6 + 2^{12}$
I think the squaring binomials/perfect square trinomials is your best general approach.
But given that the question was multiple choice, guess and check would solve this particular problem quickly ;)
answered Jan 31 at 15:10
Kurt SchwandaKurt Schwanda
40410
$endgroup$
add a comment |
$begingroup$
One way is to notice that
$28+101+2^n=128+1+2^n=2^7+1+2^n=2^n+2*2^6+1$.
Therefore, if we let $n=12$ we have
$2^{12}+2*2^6+1=(2^6)^2+2*2^6+1=(2^6+1)^2$
Which is a perfect square, as desired.
This relies on noticing this specific pattern, so doesn't lend itself to a general method. One could always guess and check.
answered Jan 31 at 15:09
Josh B.Josh B.
2,55511425
$endgroup$
add a comment |
$begingroup$
Write:$$28 + 101 + 2^n =k^2$$
For $n$ odd the expresion $$28 + 101 + 2^nequiv 2pmod 3$$ so in this case there is no solution (since $k^2equiv 0,1pmod 3$)
So $n= 2m$ and we have $$129 = k^2-2^{2m} = (k-2^m)(k+2^m)$$
Since $k-2^m<k+2^m$ we have only this possibiltes
$129 = k+2^m$ and $1=k-2^m$ ($m=6$)
or
$43 = k+2^m$ and $3=k-2^m$ (no solution)
answered Jan 31 at 15:15
Maria MazurMaria Mazur
50k1361124
$endgroup$
add a comment |
$begingroup$
Let us assume $n=2k$ and $129+2^{2k}=p^2$
Then we get:
$$p^2-(2^k)^2=129$$
$$(p-2^k)(p+2^k)=129=1 times 129=3 times 43 $$
So we get:
$$p-2^k=1$$
$$p+2^k=129$$
Solving we get
$$2^k=64$$ $implies$ $k=6$
hence $$n=12$$
The other way of resolving will not give you integer solution of $k$
answered Jan 31 at 15:34
Umesh shankarUmesh shankar
3,09231220
$endgroup$
add a comment |
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4 Answers
4
active
oldest
votes
4 Answers
4
active
oldest
votes
active
oldest
votes
active
oldest
votes
$begingroup$
You can move terms around to try to get things to look like a perfect square trinomial:
$(a + b)^2 = a^2 + 2ab + b^2$
Specifically, the special case where $a$ or $b$ is 1:
$(1 + b)^2 = 1 + 2 b + b^2$
Start by combining the constant terms, and notice that 128 is a power of 2:
$28 + 101 + 2^n = 1 + 128 + 2^n = 1 + 2 cdot 2^6 + 2^n$
We can see that $(1 + 2^6)^2 = 1 + 2 cdot 2^6 + (2^6)^2 = 1 + 2 cdot 2^6 + 2^{12}$
I think the squaring binomials/perfect square trinomials is your best general approach.
But given that the question was multiple choice, guess and check would solve this particular problem quickly ;)
answered Jan 31 at 15:10
Kurt SchwandaKurt Schwanda
40410
$endgroup$
add a comment |
$begingroup$
You can move terms around to try to get things to look like a perfect square trinomial:
$(a + b)^2 = a^2 + 2ab + b^2$
Specifically, the special case where $a$ or $b$ is 1:
$(1 + b)^2 = 1 + 2 b + b^2$
Start by combining the constant terms, and notice that 128 is a power of 2:
$28 + 101 + 2^n = 1 + 128 + 2^n = 1 + 2 cdot 2^6 + 2^n$
We can see that $(1 + 2^6)^2 = 1 + 2 cdot 2^6 + (2^6)^2 = 1 + 2 cdot 2^6 + 2^{12}$
I think the squaring binomials/perfect square trinomials is your best general approach.
But given that the question was multiple choice, guess and check would solve this particular problem quickly ;)
answered Jan 31 at 15:10
Kurt SchwandaKurt Schwanda
40410
$endgroup$
add a comment |
$begingroup$
You can move terms around to try to get things to look like a perfect square trinomial:
$(a + b)^2 = a^2 + 2ab + b^2$
Specifically, the special case where $a$ or $b$ is 1:
$(1 + b)^2 = 1 + 2 b + b^2$
Start by combining the constant terms, and notice that 128 is a power of 2:
$28 + 101 + 2^n = 1 + 128 + 2^n = 1 + 2 cdot 2^6 + 2^n$
We can see that $(1 + 2^6)^2 = 1 + 2 cdot 2^6 + (2^6)^2 = 1 + 2 cdot 2^6 + 2^{12}$
I think the squaring binomials/perfect square trinomials is your best general approach.
But given that the question was multiple choice, guess and check would solve this particular problem quickly ;)
answered Jan 31 at 15:10
Kurt SchwandaKurt Schwanda
40410
$endgroup$
You can move terms around to try to get things to look like a perfect square trinomial:
$(a + b)^2 = a^2 + 2ab + b^2$
Specifically, the special case where $a$ or $b$ is 1:
$(1 + b)^2 = 1 + 2 b + b^2$
Start by combining the constant terms, and notice that 128 is a power of 2:
$28 + 101 + 2^n = 1 + 128 + 2^n = 1 + 2 cdot 2^6 + 2^n$
We can see that $(1 + 2^6)^2 = 1 + 2 cdot 2^6 + (2^6)^2 = 1 + 2 cdot 2^6 + 2^{12}$
I think the squaring binomials/perfect square trinomials is your best general approach.
But given that the question was multiple choice, guess and check would solve this particular problem quickly ;)
answered Jan 31 at 15:10
Kurt SchwandaKurt Schwanda
40410
answered Jan 31 at 15:10
Kurt SchwandaKurt Schwanda
40410
answered Jan 31 at 15:10
Kurt SchwandaKurt Schwanda
40410
answered Jan 31 at 15:10
Kurt SchwandaKurt Schwanda
40410
40410
add a comment |
add a comment |
$begingroup$
One way is to notice that
$28+101+2^n=128+1+2^n=2^7+1+2^n=2^n+2*2^6+1$.
Therefore, if we let $n=12$ we have
$2^{12}+2*2^6+1=(2^6)^2+2*2^6+1=(2^6+1)^2$
Which is a perfect square, as desired.
This relies on noticing this specific pattern, so doesn't lend itself to a general method. One could always guess and check.
answered Jan 31 at 15:09
Josh B.Josh B.
2,55511425
$endgroup$
add a comment |
$begingroup$
One way is to notice that
$28+101+2^n=128+1+2^n=2^7+1+2^n=2^n+2*2^6+1$.
Therefore, if we let $n=12$ we have
$2^{12}+2*2^6+1=(2^6)^2+2*2^6+1=(2^6+1)^2$
Which is a perfect square, as desired.
This relies on noticing this specific pattern, so doesn't lend itself to a general method. One could always guess and check.
answered Jan 31 at 15:09
Josh B.Josh B.
2,55511425
$endgroup$
add a comment |
$begingroup$
One way is to notice that
$28+101+2^n=128+1+2^n=2^7+1+2^n=2^n+2*2^6+1$.
Therefore, if we let $n=12$ we have
$2^{12}+2*2^6+1=(2^6)^2+2*2^6+1=(2^6+1)^2$
Which is a perfect square, as desired.
This relies on noticing this specific pattern, so doesn't lend itself to a general method. One could always guess and check.
answered Jan 31 at 15:09
Josh B.Josh B.
2,55511425
$endgroup$
One way is to notice that
$28+101+2^n=128+1+2^n=2^7+1+2^n=2^n+2*2^6+1$.
Therefore, if we let $n=12$ we have
$2^{12}+2*2^6+1=(2^6)^2+2*2^6+1=(2^6+1)^2$
Which is a perfect square, as desired.
This relies on noticing this specific pattern, so doesn't lend itself to a general method. One could always guess and check.
answered Jan 31 at 15:09
Josh B.Josh B.
2,55511425
answered Jan 31 at 15:09
Josh B.Josh B.
2,55511425
answered Jan 31 at 15:09
Josh B.Josh B.
2,55511425
answered Jan 31 at 15:09
Josh B.Josh B.
2,55511425
2,55511425
add a comment |
add a comment |
$begingroup$
Write:$$28 + 101 + 2^n =k^2$$
For $n$ odd the expresion $$28 + 101 + 2^nequiv 2pmod 3$$ so in this case there is no solution (since $k^2equiv 0,1pmod 3$)
So $n= 2m$ and we have $$129 = k^2-2^{2m} = (k-2^m)(k+2^m)$$
Since $k-2^m<k+2^m$ we have only this possibiltes
$129 = k+2^m$ and $1=k-2^m$ ($m=6$)
or
$43 = k+2^m$ and $3=k-2^m$ (no solution)
answered Jan 31 at 15:15
Maria MazurMaria Mazur
50k1361124
$endgroup$
add a comment |
$begingroup$
Write:$$28 + 101 + 2^n =k^2$$
For $n$ odd the expresion $$28 + 101 + 2^nequiv 2pmod 3$$ so in this case there is no solution (since $k^2equiv 0,1pmod 3$)
So $n= 2m$ and we have $$129 = k^2-2^{2m} = (k-2^m)(k+2^m)$$
Since $k-2^m<k+2^m$ we have only this possibiltes
$129 = k+2^m$ and $1=k-2^m$ ($m=6$)
or
$43 = k+2^m$ and $3=k-2^m$ (no solution)
answered Jan 31 at 15:15
Maria MazurMaria Mazur
50k1361124
$endgroup$
add a comment |
$begingroup$
Write:$$28 + 101 + 2^n =k^2$$
For $n$ odd the expresion $$28 + 101 + 2^nequiv 2pmod 3$$ so in this case there is no solution (since $k^2equiv 0,1pmod 3$)
So $n= 2m$ and we have $$129 = k^2-2^{2m} = (k-2^m)(k+2^m)$$
Since $k-2^m<k+2^m$ we have only this possibiltes
$129 = k+2^m$ and $1=k-2^m$ ($m=6$)
or
$43 = k+2^m$ and $3=k-2^m$ (no solution)
answered Jan 31 at 15:15
Maria MazurMaria Mazur
50k1361124
$endgroup$
Write:$$28 + 101 + 2^n =k^2$$
For $n$ odd the expresion $$28 + 101 + 2^nequiv 2pmod 3$$ so in this case there is no solution (since $k^2equiv 0,1pmod 3$)
So $n= 2m$ and we have $$129 = k^2-2^{2m} = (k-2^m)(k+2^m)$$
Since $k-2^m<k+2^m$ we have only this possibiltes
$129 = k+2^m$ and $1=k-2^m$ ($m=6$)
or
$43 = k+2^m$ and $3=k-2^m$ (no solution)
answered Jan 31 at 15:15
Maria MazurMaria Mazur
50k1361124
answered Jan 31 at 15:15
Maria MazurMaria Mazur
50k1361124
answered Jan 31 at 15:15
Maria MazurMaria Mazur
50k1361124
answered Jan 31 at 15:15
Maria MazurMaria Mazur
50k1361124
50k1361124
add a comment |
add a comment |
$begingroup$
Let us assume $n=2k$ and $129+2^{2k}=p^2$
Then we get:
$$p^2-(2^k)^2=129$$
$$(p-2^k)(p+2^k)=129=1 times 129=3 times 43 $$
So we get:
$$p-2^k=1$$
$$p+2^k=129$$
Solving we get
$$2^k=64$$ $implies$ $k=6$
hence $$n=12$$
The other way of resolving will not give you integer solution of $k$
answered Jan 31 at 15:34
Umesh shankarUmesh shankar
3,09231220
$endgroup$
add a comment |
$begingroup$
Let us assume $n=2k$ and $129+2^{2k}=p^2$
Then we get:
$$p^2-(2^k)^2=129$$
$$(p-2^k)(p+2^k)=129=1 times 129=3 times 43 $$
So we get:
$$p-2^k=1$$
$$p+2^k=129$$
Solving we get
$$2^k=64$$ $implies$ $k=6$
hence $$n=12$$
The other way of resolving will not give you integer solution of $k$
answered Jan 31 at 15:34
Umesh shankarUmesh shankar
3,09231220
$endgroup$
add a comment |
$begingroup$
Let us assume $n=2k$ and $129+2^{2k}=p^2$
Then we get:
$$p^2-(2^k)^2=129$$
$$(p-2^k)(p+2^k)=129=1 times 129=3 times 43 $$
So we get:
$$p-2^k=1$$
$$p+2^k=129$$
Solving we get
$$2^k=64$$ $implies$ $k=6$
hence $$n=12$$
The other way of resolving will not give you integer solution of $k$
answered Jan 31 at 15:34
Umesh shankarUmesh shankar
3,09231220
$endgroup$
Let us assume $n=2k$ and $129+2^{2k}=p^2$
Then we get:
$$p^2-(2^k)^2=129$$
$$(p-2^k)(p+2^k)=129=1 times 129=3 times 43 $$
So we get:
$$p-2^k=1$$
$$p+2^k=129$$
Solving we get
$$2^k=64$$ $implies$ $k=6$
hence $$n=12$$
The other way of resolving will not give you integer solution of $k$
answered Jan 31 at 15:34
Umesh shankarUmesh shankar
3,09231220
answered Jan 31 at 15:34
Umesh shankarUmesh shankar
3,09231220
answered Jan 31 at 15:34
Umesh shankarUmesh shankar
3,09231220
answered Jan 31 at 15:34
Umesh shankarUmesh shankar
3,09231220
3,09231220
add a comment |
add a comment |
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$begingroup$
Try $$n=12$$ to get a square number
$endgroup$
– Dr. Sonnhard Graubner
Jan 31 at 15:09