How To solve This Perfect Square Word Problem
Clash Royale CLAN TAG#URR8PPP
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Here's a problem about perfect squares and it's very hard for me. I tried to solve but I got stuck.
Last year, the town of Whipple had a population that was a perfect square. Last month, 100 enlightened people moved to Whipple, making the population one more than a perfect square. Next month, 100 more people will move to Whipple, making the population a perfect square again. What was the original population of Whipple?
Here's what I did:
Let the population last year be n, so n = x^2 and x = √n
Last month: n + 100 = x^2 + 1
Next Month: n + 200 = x^2 ...
and i Got stuck there. I don't know where I am going ... Your help is appreciated
algebra-precalculus square-numbers
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add a comment |
$begingroup$
Here's a problem about perfect squares and it's very hard for me. I tried to solve but I got stuck.
Last year, the town of Whipple had a population that was a perfect square. Last month, 100 enlightened people moved to Whipple, making the population one more than a perfect square. Next month, 100 more people will move to Whipple, making the population a perfect square again. What was the original population of Whipple?
Here's what I did:
Let the population last year be n, so n = x^2 and x = √n
Last month: n + 100 = x^2 + 1
Next Month: n + 200 = x^2 ...
and i Got stuck there. I don't know where I am going ... Your help is appreciated
algebra-precalculus square-numbers
$endgroup$
add a comment |
$begingroup$
Here's a problem about perfect squares and it's very hard for me. I tried to solve but I got stuck.
Last year, the town of Whipple had a population that was a perfect square. Last month, 100 enlightened people moved to Whipple, making the population one more than a perfect square. Next month, 100 more people will move to Whipple, making the population a perfect square again. What was the original population of Whipple?
Here's what I did:
Let the population last year be n, so n = x^2 and x = √n
Last month: n + 100 = x^2 + 1
Next Month: n + 200 = x^2 ...
and i Got stuck there. I don't know where I am going ... Your help is appreciated
algebra-precalculus square-numbers
$endgroup$
Here's a problem about perfect squares and it's very hard for me. I tried to solve but I got stuck.
Last year, the town of Whipple had a population that was a perfect square. Last month, 100 enlightened people moved to Whipple, making the population one more than a perfect square. Next month, 100 more people will move to Whipple, making the population a perfect square again. What was the original population of Whipple?
Here's what I did:
Let the population last year be n, so n = x^2 and x = √n
Last month: n + 100 = x^2 + 1
Next Month: n + 200 = x^2 ...
and i Got stuck there. I don't know where I am going ... Your help is appreciated
algebra-precalculus square-numbers
algebra-precalculus square-numbers
asked Jan 29 at 23:38
harpey1111harpey1111
555
555
add a comment |
add a comment |
4 Answers
4
active
oldest
votes
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I like where Will Jagy starts.
$$x^2+99 = y^2\
x^2 + 200 = z^2$$
to continue I would subtract one from the other
$$z^2 - y^2 = 101\(z+y)(z-y) = 101$$
$101$ is prime
$$z+y = 101\z-y = 1\z = 51\y = 50\x = sqrt51^2 - 200 = 49$$
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good...........
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– Will Jagy
Jan 30 at 0:43
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on this part, I just don't understand how you got the z - y = 100 and why you need to subtract the two equations. :(
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– harpey1111
Jan 30 at 22:34
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and maybe that 50^2 is supposed tobe 51^2 right?
$endgroup$
– harpey1111
Jan 30 at 22:45
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@harpey1111 I subtracted one from the other, for a couple of reasons. One, it eliminates the x^2 term. Two, it creates a prime number to work with. And yes, $51^2 - 200 = 49^2$ thanks for pointing that out.
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– Doug M
Jan 30 at 23:08
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thank you so much for your help. I got the problem right. have a wonderful night :)
$endgroup$
– harpey1111
Jan 30 at 23:34
add a comment |
$begingroup$
Be careful with your variable names; it will clarify things for you. You've used $x$ to mean three different things!
Let $n=x^2$; then $n+200=y^2$, so that $x^2+200 = y^2$. Rewriting gives $y^2-x^2=(y-x)(y+x)=200$. Can you make progress from there?
$endgroup$
$begingroup$
I kind of get it now but I still don't know how to use it to solve the problem because the "Last month, 100 enlightened people moved to Whipple, making the population one more than a perfect square." makes me really confused. I'm so sorry.
$endgroup$
– harpey1111
Jan 29 at 23:57
$begingroup$
thank you. I got the problem right . I really appreciate your help. and I'll take note of your advice. it's really helpful.
$endgroup$
– harpey1111
Jan 30 at 23:38
$begingroup$
You're welcome. You should accept, by clicking the green checkmark, whichever of the several great answers you like best.
$endgroup$
– rogerl
Jan 31 at 14:06
add a comment |
$begingroup$
You have a square $x^2$ and you know that $x^2+99=y^2$ and $y^2+101=z^2$.
It is well known that the sum of the first $n$ odd numbers equals $n^2$, meaning that adding sequential odd numbers to a smaller square yields the squares of the next larger numbers, i.e. if the first $n$ odd numbers sum to $n^2$ then the first $n+1$ odd numbers sum to $(n+1)^2$, etc.
Now notice that $99$ and $101$ are sequential odd numbers. A little manipulation reveals that $99$ is the $50$th odd number ($99=(2cdot 50)-1)$) and $101$ is the $51$st odd number ($101=(2cdot 51)-1)$). The starting square in this scenario would be the square which is the sum of the first $49$ odd numbers. So the squares that solve the problem are $49^2$, $50^2$, and $51^2$.
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thank you for spending some of your time explaining this. It helped me. I really appreciate it.
$endgroup$
– harpey1111
Jan 30 at 23:39
add a comment |
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$$ x^2 + 99 = y^2 $$
$$ x^2 + 200 = z^2 $$
$$ (y+x)(y-x) = 99 = 99 cdot 1 = 33 cdot 3 = 11 cdot 9$$
The choices for $x$ are
$$ frac99-12 , ; ; ; frac33-32 , ; ; ; frac11-92 , ; ; ; $$
Then confirm that $x^2 + 99$ is really a square, and check whether $x^2 + 200$ is a square, for the three possible $x$ values.
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how did you get "divided by 2"?
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– harpey1111
Jan 30 at 0:32
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@harpey1111 what are $y$ and $x$ when $y+x=99$ and $y-x=1?$
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– Will Jagy
Jan 30 at 0:37
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Thank you so much. I got the problem right. thanks a lot. Have a wonderful night :)
$endgroup$
– harpey1111
Jan 30 at 23:40
add a comment |
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4 Answers
4
active
oldest
votes
4 Answers
4
active
oldest
votes
active
oldest
votes
active
oldest
votes
$begingroup$
I like where Will Jagy starts.
$$x^2+99 = y^2\
x^2 + 200 = z^2$$
to continue I would subtract one from the other
$$z^2 - y^2 = 101\(z+y)(z-y) = 101$$
$101$ is prime
$$z+y = 101\z-y = 1\z = 51\y = 50\x = sqrt51^2 - 200 = 49$$
$endgroup$
$begingroup$
good...........
$endgroup$
– Will Jagy
Jan 30 at 0:43
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on this part, I just don't understand how you got the z - y = 100 and why you need to subtract the two equations. :(
$endgroup$
– harpey1111
Jan 30 at 22:34
$begingroup$
and maybe that 50^2 is supposed tobe 51^2 right?
$endgroup$
– harpey1111
Jan 30 at 22:45
$begingroup$
@harpey1111 I subtracted one from the other, for a couple of reasons. One, it eliminates the x^2 term. Two, it creates a prime number to work with. And yes, $51^2 - 200 = 49^2$ thanks for pointing that out.
$endgroup$
– Doug M
Jan 30 at 23:08
$begingroup$
thank you so much for your help. I got the problem right. have a wonderful night :)
$endgroup$
– harpey1111
Jan 30 at 23:34
add a comment |
$begingroup$
I like where Will Jagy starts.
$$x^2+99 = y^2\
x^2 + 200 = z^2$$
to continue I would subtract one from the other
$$z^2 - y^2 = 101\(z+y)(z-y) = 101$$
$101$ is prime
$$z+y = 101\z-y = 1\z = 51\y = 50\x = sqrt51^2 - 200 = 49$$
$endgroup$
$begingroup$
good...........
$endgroup$
– Will Jagy
Jan 30 at 0:43
$begingroup$
on this part, I just don't understand how you got the z - y = 100 and why you need to subtract the two equations. :(
$endgroup$
– harpey1111
Jan 30 at 22:34
$begingroup$
and maybe that 50^2 is supposed tobe 51^2 right?
$endgroup$
– harpey1111
Jan 30 at 22:45
$begingroup$
@harpey1111 I subtracted one from the other, for a couple of reasons. One, it eliminates the x^2 term. Two, it creates a prime number to work with. And yes, $51^2 - 200 = 49^2$ thanks for pointing that out.
$endgroup$
– Doug M
Jan 30 at 23:08
$begingroup$
thank you so much for your help. I got the problem right. have a wonderful night :)
$endgroup$
– harpey1111
Jan 30 at 23:34
add a comment |
$begingroup$
I like where Will Jagy starts.
$$x^2+99 = y^2\
x^2 + 200 = z^2$$
to continue I would subtract one from the other
$$z^2 - y^2 = 101\(z+y)(z-y) = 101$$
$101$ is prime
$$z+y = 101\z-y = 1\z = 51\y = 50\x = sqrt51^2 - 200 = 49$$
$endgroup$
I like where Will Jagy starts.
$$x^2+99 = y^2\
x^2 + 200 = z^2$$
to continue I would subtract one from the other
$$z^2 - y^2 = 101\(z+y)(z-y) = 101$$
$101$ is prime
$$z+y = 101\z-y = 1\z = 51\y = 50\x = sqrt51^2 - 200 = 49$$
edited Jan 30 at 23:05
answered Jan 30 at 0:37
Doug MDoug M
45.2k31854
45.2k31854
$begingroup$
good...........
$endgroup$
– Will Jagy
Jan 30 at 0:43
$begingroup$
on this part, I just don't understand how you got the z - y = 100 and why you need to subtract the two equations. :(
$endgroup$
– harpey1111
Jan 30 at 22:34
$begingroup$
and maybe that 50^2 is supposed tobe 51^2 right?
$endgroup$
– harpey1111
Jan 30 at 22:45
$begingroup$
@harpey1111 I subtracted one from the other, for a couple of reasons. One, it eliminates the x^2 term. Two, it creates a prime number to work with. And yes, $51^2 - 200 = 49^2$ thanks for pointing that out.
$endgroup$
– Doug M
Jan 30 at 23:08
$begingroup$
thank you so much for your help. I got the problem right. have a wonderful night :)
$endgroup$
– harpey1111
Jan 30 at 23:34
add a comment |
$begingroup$
good...........
$endgroup$
– Will Jagy
Jan 30 at 0:43
$begingroup$
on this part, I just don't understand how you got the z - y = 100 and why you need to subtract the two equations. :(
$endgroup$
– harpey1111
Jan 30 at 22:34
$begingroup$
and maybe that 50^2 is supposed tobe 51^2 right?
$endgroup$
– harpey1111
Jan 30 at 22:45
$begingroup$
@harpey1111 I subtracted one from the other, for a couple of reasons. One, it eliminates the x^2 term. Two, it creates a prime number to work with. And yes, $51^2 - 200 = 49^2$ thanks for pointing that out.
$endgroup$
– Doug M
Jan 30 at 23:08
$begingroup$
thank you so much for your help. I got the problem right. have a wonderful night :)
$endgroup$
– harpey1111
Jan 30 at 23:34
$begingroup$
good...........
$endgroup$
– Will Jagy
Jan 30 at 0:43
$begingroup$
good...........
$endgroup$
– Will Jagy
Jan 30 at 0:43
$begingroup$
on this part, I just don't understand how you got the z - y = 100 and why you need to subtract the two equations. :(
$endgroup$
– harpey1111
Jan 30 at 22:34
$begingroup$
on this part, I just don't understand how you got the z - y = 100 and why you need to subtract the two equations. :(
$endgroup$
– harpey1111
Jan 30 at 22:34
$begingroup$
and maybe that 50^2 is supposed tobe 51^2 right?
$endgroup$
– harpey1111
Jan 30 at 22:45
$begingroup$
and maybe that 50^2 is supposed tobe 51^2 right?
$endgroup$
– harpey1111
Jan 30 at 22:45
$begingroup$
@harpey1111 I subtracted one from the other, for a couple of reasons. One, it eliminates the x^2 term. Two, it creates a prime number to work with. And yes, $51^2 - 200 = 49^2$ thanks for pointing that out.
$endgroup$
– Doug M
Jan 30 at 23:08
$begingroup$
@harpey1111 I subtracted one from the other, for a couple of reasons. One, it eliminates the x^2 term. Two, it creates a prime number to work with. And yes, $51^2 - 200 = 49^2$ thanks for pointing that out.
$endgroup$
– Doug M
Jan 30 at 23:08
$begingroup$
thank you so much for your help. I got the problem right. have a wonderful night :)
$endgroup$
– harpey1111
Jan 30 at 23:34
$begingroup$
thank you so much for your help. I got the problem right. have a wonderful night :)
$endgroup$
– harpey1111
Jan 30 at 23:34
add a comment |
$begingroup$
Be careful with your variable names; it will clarify things for you. You've used $x$ to mean three different things!
Let $n=x^2$; then $n+200=y^2$, so that $x^2+200 = y^2$. Rewriting gives $y^2-x^2=(y-x)(y+x)=200$. Can you make progress from there?
$endgroup$
$begingroup$
I kind of get it now but I still don't know how to use it to solve the problem because the "Last month, 100 enlightened people moved to Whipple, making the population one more than a perfect square." makes me really confused. I'm so sorry.
$endgroup$
– harpey1111
Jan 29 at 23:57
$begingroup$
thank you. I got the problem right . I really appreciate your help. and I'll take note of your advice. it's really helpful.
$endgroup$
– harpey1111
Jan 30 at 23:38
$begingroup$
You're welcome. You should accept, by clicking the green checkmark, whichever of the several great answers you like best.
$endgroup$
– rogerl
Jan 31 at 14:06
add a comment |
$begingroup$
Be careful with your variable names; it will clarify things for you. You've used $x$ to mean three different things!
Let $n=x^2$; then $n+200=y^2$, so that $x^2+200 = y^2$. Rewriting gives $y^2-x^2=(y-x)(y+x)=200$. Can you make progress from there?
$endgroup$
$begingroup$
I kind of get it now but I still don't know how to use it to solve the problem because the "Last month, 100 enlightened people moved to Whipple, making the population one more than a perfect square." makes me really confused. I'm so sorry.
$endgroup$
– harpey1111
Jan 29 at 23:57
$begingroup$
thank you. I got the problem right . I really appreciate your help. and I'll take note of your advice. it's really helpful.
$endgroup$
– harpey1111
Jan 30 at 23:38
$begingroup$
You're welcome. You should accept, by clicking the green checkmark, whichever of the several great answers you like best.
$endgroup$
– rogerl
Jan 31 at 14:06
add a comment |
$begingroup$
Be careful with your variable names; it will clarify things for you. You've used $x$ to mean three different things!
Let $n=x^2$; then $n+200=y^2$, so that $x^2+200 = y^2$. Rewriting gives $y^2-x^2=(y-x)(y+x)=200$. Can you make progress from there?
$endgroup$
Be careful with your variable names; it will clarify things for you. You've used $x$ to mean three different things!
Let $n=x^2$; then $n+200=y^2$, so that $x^2+200 = y^2$. Rewriting gives $y^2-x^2=(y-x)(y+x)=200$. Can you make progress from there?
answered Jan 29 at 23:44
rogerlrogerl
17.9k22747
17.9k22747
$begingroup$
I kind of get it now but I still don't know how to use it to solve the problem because the "Last month, 100 enlightened people moved to Whipple, making the population one more than a perfect square." makes me really confused. I'm so sorry.
$endgroup$
– harpey1111
Jan 29 at 23:57
$begingroup$
thank you. I got the problem right . I really appreciate your help. and I'll take note of your advice. it's really helpful.
$endgroup$
– harpey1111
Jan 30 at 23:38
$begingroup$
You're welcome. You should accept, by clicking the green checkmark, whichever of the several great answers you like best.
$endgroup$
– rogerl
Jan 31 at 14:06
add a comment |
$begingroup$
I kind of get it now but I still don't know how to use it to solve the problem because the "Last month, 100 enlightened people moved to Whipple, making the population one more than a perfect square." makes me really confused. I'm so sorry.
$endgroup$
– harpey1111
Jan 29 at 23:57
$begingroup$
thank you. I got the problem right . I really appreciate your help. and I'll take note of your advice. it's really helpful.
$endgroup$
– harpey1111
Jan 30 at 23:38
$begingroup$
You're welcome. You should accept, by clicking the green checkmark, whichever of the several great answers you like best.
$endgroup$
– rogerl
Jan 31 at 14:06
$begingroup$
I kind of get it now but I still don't know how to use it to solve the problem because the "Last month, 100 enlightened people moved to Whipple, making the population one more than a perfect square." makes me really confused. I'm so sorry.
$endgroup$
– harpey1111
Jan 29 at 23:57
$begingroup$
I kind of get it now but I still don't know how to use it to solve the problem because the "Last month, 100 enlightened people moved to Whipple, making the population one more than a perfect square." makes me really confused. I'm so sorry.
$endgroup$
– harpey1111
Jan 29 at 23:57
$begingroup$
thank you. I got the problem right . I really appreciate your help. and I'll take note of your advice. it's really helpful.
$endgroup$
– harpey1111
Jan 30 at 23:38
$begingroup$
thank you. I got the problem right . I really appreciate your help. and I'll take note of your advice. it's really helpful.
$endgroup$
– harpey1111
Jan 30 at 23:38
$begingroup$
You're welcome. You should accept, by clicking the green checkmark, whichever of the several great answers you like best.
$endgroup$
– rogerl
Jan 31 at 14:06
$begingroup$
You're welcome. You should accept, by clicking the green checkmark, whichever of the several great answers you like best.
$endgroup$
– rogerl
Jan 31 at 14:06
add a comment |
$begingroup$
You have a square $x^2$ and you know that $x^2+99=y^2$ and $y^2+101=z^2$.
It is well known that the sum of the first $n$ odd numbers equals $n^2$, meaning that adding sequential odd numbers to a smaller square yields the squares of the next larger numbers, i.e. if the first $n$ odd numbers sum to $n^2$ then the first $n+1$ odd numbers sum to $(n+1)^2$, etc.
Now notice that $99$ and $101$ are sequential odd numbers. A little manipulation reveals that $99$ is the $50$th odd number ($99=(2cdot 50)-1)$) and $101$ is the $51$st odd number ($101=(2cdot 51)-1)$). The starting square in this scenario would be the square which is the sum of the first $49$ odd numbers. So the squares that solve the problem are $49^2$, $50^2$, and $51^2$.
$endgroup$
$begingroup$
thank you for spending some of your time explaining this. It helped me. I really appreciate it.
$endgroup$
– harpey1111
Jan 30 at 23:39
add a comment |
$begingroup$
You have a square $x^2$ and you know that $x^2+99=y^2$ and $y^2+101=z^2$.
It is well known that the sum of the first $n$ odd numbers equals $n^2$, meaning that adding sequential odd numbers to a smaller square yields the squares of the next larger numbers, i.e. if the first $n$ odd numbers sum to $n^2$ then the first $n+1$ odd numbers sum to $(n+1)^2$, etc.
Now notice that $99$ and $101$ are sequential odd numbers. A little manipulation reveals that $99$ is the $50$th odd number ($99=(2cdot 50)-1)$) and $101$ is the $51$st odd number ($101=(2cdot 51)-1)$). The starting square in this scenario would be the square which is the sum of the first $49$ odd numbers. So the squares that solve the problem are $49^2$, $50^2$, and $51^2$.
$endgroup$
$begingroup$
thank you for spending some of your time explaining this. It helped me. I really appreciate it.
$endgroup$
– harpey1111
Jan 30 at 23:39
add a comment |
$begingroup$
You have a square $x^2$ and you know that $x^2+99=y^2$ and $y^2+101=z^2$.
It is well known that the sum of the first $n$ odd numbers equals $n^2$, meaning that adding sequential odd numbers to a smaller square yields the squares of the next larger numbers, i.e. if the first $n$ odd numbers sum to $n^2$ then the first $n+1$ odd numbers sum to $(n+1)^2$, etc.
Now notice that $99$ and $101$ are sequential odd numbers. A little manipulation reveals that $99$ is the $50$th odd number ($99=(2cdot 50)-1)$) and $101$ is the $51$st odd number ($101=(2cdot 51)-1)$). The starting square in this scenario would be the square which is the sum of the first $49$ odd numbers. So the squares that solve the problem are $49^2$, $50^2$, and $51^2$.
$endgroup$
You have a square $x^2$ and you know that $x^2+99=y^2$ and $y^2+101=z^2$.
It is well known that the sum of the first $n$ odd numbers equals $n^2$, meaning that adding sequential odd numbers to a smaller square yields the squares of the next larger numbers, i.e. if the first $n$ odd numbers sum to $n^2$ then the first $n+1$ odd numbers sum to $(n+1)^2$, etc.
Now notice that $99$ and $101$ are sequential odd numbers. A little manipulation reveals that $99$ is the $50$th odd number ($99=(2cdot 50)-1)$) and $101$ is the $51$st odd number ($101=(2cdot 51)-1)$). The starting square in this scenario would be the square which is the sum of the first $49$ odd numbers. So the squares that solve the problem are $49^2$, $50^2$, and $51^2$.
answered Jan 30 at 2:48
Keith BackmanKeith Backman
1,3681812
1,3681812
$begingroup$
thank you for spending some of your time explaining this. It helped me. I really appreciate it.
$endgroup$
– harpey1111
Jan 30 at 23:39
add a comment |
$begingroup$
thank you for spending some of your time explaining this. It helped me. I really appreciate it.
$endgroup$
– harpey1111
Jan 30 at 23:39
$begingroup$
thank you for spending some of your time explaining this. It helped me. I really appreciate it.
$endgroup$
– harpey1111
Jan 30 at 23:39
$begingroup$
thank you for spending some of your time explaining this. It helped me. I really appreciate it.
$endgroup$
– harpey1111
Jan 30 at 23:39
add a comment |
$begingroup$
$$ x^2 + 99 = y^2 $$
$$ x^2 + 200 = z^2 $$
$$ (y+x)(y-x) = 99 = 99 cdot 1 = 33 cdot 3 = 11 cdot 9$$
The choices for $x$ are
$$ frac99-12 , ; ; ; frac33-32 , ; ; ; frac11-92 , ; ; ; $$
Then confirm that $x^2 + 99$ is really a square, and check whether $x^2 + 200$ is a square, for the three possible $x$ values.
$endgroup$
$begingroup$
how did you get "divided by 2"?
$endgroup$
– harpey1111
Jan 30 at 0:32
$begingroup$
@harpey1111 what are $y$ and $x$ when $y+x=99$ and $y-x=1?$
$endgroup$
– Will Jagy
Jan 30 at 0:37
$begingroup$
Thank you so much. I got the problem right. thanks a lot. Have a wonderful night :)
$endgroup$
– harpey1111
Jan 30 at 23:40
add a comment |
$begingroup$
$$ x^2 + 99 = y^2 $$
$$ x^2 + 200 = z^2 $$
$$ (y+x)(y-x) = 99 = 99 cdot 1 = 33 cdot 3 = 11 cdot 9$$
The choices for $x$ are
$$ frac99-12 , ; ; ; frac33-32 , ; ; ; frac11-92 , ; ; ; $$
Then confirm that $x^2 + 99$ is really a square, and check whether $x^2 + 200$ is a square, for the three possible $x$ values.
$endgroup$
$begingroup$
how did you get "divided by 2"?
$endgroup$
– harpey1111
Jan 30 at 0:32
$begingroup$
@harpey1111 what are $y$ and $x$ when $y+x=99$ and $y-x=1?$
$endgroup$
– Will Jagy
Jan 30 at 0:37
$begingroup$
Thank you so much. I got the problem right. thanks a lot. Have a wonderful night :)
$endgroup$
– harpey1111
Jan 30 at 23:40
add a comment |
$begingroup$
$$ x^2 + 99 = y^2 $$
$$ x^2 + 200 = z^2 $$
$$ (y+x)(y-x) = 99 = 99 cdot 1 = 33 cdot 3 = 11 cdot 9$$
The choices for $x$ are
$$ frac99-12 , ; ; ; frac33-32 , ; ; ; frac11-92 , ; ; ; $$
Then confirm that $x^2 + 99$ is really a square, and check whether $x^2 + 200$ is a square, for the three possible $x$ values.
$endgroup$
$$ x^2 + 99 = y^2 $$
$$ x^2 + 200 = z^2 $$
$$ (y+x)(y-x) = 99 = 99 cdot 1 = 33 cdot 3 = 11 cdot 9$$
The choices for $x$ are
$$ frac99-12 , ; ; ; frac33-32 , ; ; ; frac11-92 , ; ; ; $$
Then confirm that $x^2 + 99$ is really a square, and check whether $x^2 + 200$ is a square, for the three possible $x$ values.
edited Jan 30 at 0:26
answered Jan 30 at 0:20
Will JagyWill Jagy
103k5102200
103k5102200
$begingroup$
how did you get "divided by 2"?
$endgroup$
– harpey1111
Jan 30 at 0:32
$begingroup$
@harpey1111 what are $y$ and $x$ when $y+x=99$ and $y-x=1?$
$endgroup$
– Will Jagy
Jan 30 at 0:37
$begingroup$
Thank you so much. I got the problem right. thanks a lot. Have a wonderful night :)
$endgroup$
– harpey1111
Jan 30 at 23:40
add a comment |
$begingroup$
how did you get "divided by 2"?
$endgroup$
– harpey1111
Jan 30 at 0:32
$begingroup$
@harpey1111 what are $y$ and $x$ when $y+x=99$ and $y-x=1?$
$endgroup$
– Will Jagy
Jan 30 at 0:37
$begingroup$
Thank you so much. I got the problem right. thanks a lot. Have a wonderful night :)
$endgroup$
– harpey1111
Jan 30 at 23:40
$begingroup$
how did you get "divided by 2"?
$endgroup$
– harpey1111
Jan 30 at 0:32
$begingroup$
how did you get "divided by 2"?
$endgroup$
– harpey1111
Jan 30 at 0:32
$begingroup$
@harpey1111 what are $y$ and $x$ when $y+x=99$ and $y-x=1?$
$endgroup$
– Will Jagy
Jan 30 at 0:37
$begingroup$
@harpey1111 what are $y$ and $x$ when $y+x=99$ and $y-x=1?$
$endgroup$
– Will Jagy
Jan 30 at 0:37
$begingroup$
Thank you so much. I got the problem right. thanks a lot. Have a wonderful night :)
$endgroup$
– harpey1111
Jan 30 at 23:40
$begingroup$
Thank you so much. I got the problem right. thanks a lot. Have a wonderful night :)
$endgroup$
– harpey1111
Jan 30 at 23:40
add a comment |
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