The difference between two perfect squares Help!!! [closed]

0

$begingroup$

The difference between two perfect squares is $ 133 $. What is the smallest possible sum of the two perfect squares?

I would like a step by step explanation.

Thank you.

algebra-precalculus

share|cite|improve this question

edited Jan 21 at 3:09

user549397

1,5061418

asked Jan 21 at 1:50

user636736user636736

1

$endgroup$

closed as off-topic by rschwieb, max_zorn, Leucippus, user91500, Shailesh Jan 21 at 6:53

This question appears to be off-topic. The users who voted to close gave this specific reason:

• "This question is missing context or other details: Please provide additional context, which ideally explains why the question is relevant to you and our community. Some forms of context include: background and motivation, relevant definitions, source, possible strategies, your current progress, why the question is interesting or important, etc." – rschwieb, max_zorn, Leucippus, user91500, Shailesh

If this question can be reworded to fit the rules in the help center, please edit the question.

• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  Hint: $n^2-m^2=(n+m)times(n-m)$
  $endgroup$
  – J. W. Tanner
  Jan 21 at 1:53
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  Do you know what this means about factoring $133$?
  $endgroup$
  – hardmath
  Jan 21 at 1:53
  

  
  
  
  


• 
  
  
  

  
  2
  

  
  
  
  
  
  

  
  $begingroup$
  Welcome to stackexchange. Although you would like a complete solution you are more likely to get help if you show us what you tried. Here's a hint. Factor $133$ and remember that $a^2-b^2 = (a-b)(a+b).
  $endgroup$
  – Ethan Bolker
  Jan 21 at 1:54
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  Any time you hear of the difference of squares you should think of the factorization J. W. Tanner gave. It comes up all the time.
  $endgroup$
  – Ross Millikan
  Jan 21 at 2:10
  

  
  
  
  

add a comment | 

0

$begingroup$

The difference between two perfect squares is $ 133 $. What is the smallest possible sum of the two perfect squares?

I would like a step by step explanation.

Thank you.

algebra-precalculus

share|cite|improve this question

edited Jan 21 at 3:09

user549397

1,5061418

asked Jan 21 at 1:50

user636736user636736

1

$endgroup$

closed as off-topic by rschwieb, max_zorn, Leucippus, user91500, Shailesh Jan 21 at 6:53

This question appears to be off-topic. The users who voted to close gave this specific reason:

• "This question is missing context or other details: Please provide additional context, which ideally explains why the question is relevant to you and our community. Some forms of context include: background and motivation, relevant definitions, source, possible strategies, your current progress, why the question is interesting or important, etc." – rschwieb, max_zorn, Leucippus, user91500, Shailesh

If this question can be reworded to fit the rules in the help center, please edit the question.

• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  Hint: $n^2-m^2=(n+m)times(n-m)$
  $endgroup$
  – J. W. Tanner
  Jan 21 at 1:53
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  Do you know what this means about factoring $133$?
  $endgroup$
  – hardmath
  Jan 21 at 1:53
  

  
  
  
  


• 
  
  
  

  
  2
  

  
  
  
  
  
  

  
  $begingroup$
  Welcome to stackexchange. Although you would like a complete solution you are more likely to get help if you show us what you tried. Here's a hint. Factor $133$ and remember that $a^2-b^2 = (a-b)(a+b).
  $endgroup$
  – Ethan Bolker
  Jan 21 at 1:54
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  Any time you hear of the difference of squares you should think of the factorization J. W. Tanner gave. It comes up all the time.
  $endgroup$
  – Ross Millikan
  Jan 21 at 2:10
  

  
  
  
  

add a comment | 

0

0

0

$begingroup$

The difference between two perfect squares is $ 133 $. What is the smallest possible sum of the two perfect squares?

I would like a step by step explanation.

Thank you.

algebra-precalculus

share|cite|improve this question

edited Jan 21 at 3:09

user549397

1,5061418

asked Jan 21 at 1:50

user636736user636736

1

$endgroup$

The difference between two perfect squares is $ 133 $. What is the smallest possible sum of the two perfect squares?

I would like a step by step explanation.

Thank you.

algebra-precalculus

algebra-precalculus

share|cite|improve this question

edited Jan 21 at 3:09

user549397

1,5061418

asked Jan 21 at 1:50

user636736user636736

1

share|cite|improve this question

edited Jan 21 at 3:09

user549397

1,5061418

asked Jan 21 at 1:50

user636736user636736

1

share|cite|improve this question

share|cite|improve this question

edited Jan 21 at 3:09

user549397

1,5061418

edited Jan 21 at 3:09

user549397

1,5061418

edited Jan 21 at 3:09

user549397

1,5061418

1,5061418

asked Jan 21 at 1:50

user636736user636736

1

asked Jan 21 at 1:50

user636736user636736

1

asked Jan 21 at 1:50

user636736user636736

1

1

closed as off-topic by rschwieb, max_zorn, Leucippus, user91500, Shailesh Jan 21 at 6:53

This question appears to be off-topic. The users who voted to close gave this specific reason:

• "This question is missing context or other details: Please provide additional context, which ideally explains why the question is relevant to you and our community. Some forms of context include: background and motivation, relevant definitions, source, possible strategies, your current progress, why the question is interesting or important, etc." – rschwieb, max_zorn, Leucippus, user91500, Shailesh

If this question can be reworded to fit the rules in the help center, please edit the question.

closed as off-topic by rschwieb, max_zorn, Leucippus, user91500, Shailesh Jan 21 at 6:53

This question appears to be off-topic. The users who voted to close gave this specific reason:

• "This question is missing context or other details: Please provide additional context, which ideally explains why the question is relevant to you and our community. Some forms of context include: background and motivation, relevant definitions, source, possible strategies, your current progress, why the question is interesting or important, etc." – rschwieb, max_zorn, Leucippus, user91500, Shailesh

If this question can be reworded to fit the rules in the help center, please edit the question.

• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  Hint: $n^2-m^2=(n+m)times(n-m)$
  $endgroup$
  – J. W. Tanner
  Jan 21 at 1:53
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  Do you know what this means about factoring $133$?
  $endgroup$
  – hardmath
  Jan 21 at 1:53
  

  
  
  
  


• 
  
  
  

  
  2
  

  
  
  
  
  
  

  
  $begingroup$
  Welcome to stackexchange. Although you would like a complete solution you are more likely to get help if you show us what you tried. Here's a hint. Factor $133$ and remember that $a^2-b^2 = (a-b)(a+b).
  $endgroup$
  – Ethan Bolker
  Jan 21 at 1:54
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  Any time you hear of the difference of squares you should think of the factorization J. W. Tanner gave. It comes up all the time.
  $endgroup$
  – Ross Millikan
  Jan 21 at 2:10
  

  
  
  
  

add a comment | 

• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  Hint: $n^2-m^2=(n+m)times(n-m)$
  $endgroup$
  – J. W. Tanner
  Jan 21 at 1:53
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  Do you know what this means about factoring $133$?
  $endgroup$
  – hardmath
  Jan 21 at 1:53
  

  
  
  
  


• 
  
  
  

  
  2
  

  
  
  
  
  
  

  
  $begingroup$
  Welcome to stackexchange. Although you would like a complete solution you are more likely to get help if you show us what you tried. Here's a hint. Factor $133$ and remember that $a^2-b^2 = (a-b)(a+b).
  $endgroup$
  – Ethan Bolker
  Jan 21 at 1:54
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  Any time you hear of the difference of squares you should think of the factorization J. W. Tanner gave. It comes up all the time.
  $endgroup$
  – Ross Millikan
  Jan 21 at 2:10
  

  
  
  
  

$begingroup$
Hint: $n^2-m^2=(n+m)times(n-m)$
$endgroup$
– J. W. Tanner
Jan 21 at 1:53

$begingroup$
Hint: $n^2-m^2=(n+m)times(n-m)$
$endgroup$
– J. W. Tanner
Jan 21 at 1:53

$begingroup$
Do you know what this means about factoring $133$?
$endgroup$
– hardmath
Jan 21 at 1:53

$begingroup$
Do you know what this means about factoring $133$?
$endgroup$
– hardmath
Jan 21 at 1:53

2

2

$begingroup$
Welcome to stackexchange. Although you would like a complete solution you are more likely to get help if you show us what you tried. Here's a hint. Factor $133$ and remember that $a^2-b^2 = (a-b)(a+b).
$endgroup$
– Ethan Bolker
Jan 21 at 1:54

$begingroup$
Welcome to stackexchange. Although you would like a complete solution you are more likely to get help if you show us what you tried. Here's a hint. Factor $133$ and remember that $a^2-b^2 = (a-b)(a+b).
$endgroup$
– Ethan Bolker
Jan 21 at 1:54

$begingroup$
Any time you hear of the difference of squares you should think of the factorization J. W. Tanner gave. It comes up all the time.
$endgroup$
– Ross Millikan
Jan 21 at 2:10

$begingroup$
Any time you hear of the difference of squares you should think of the factorization J. W. Tanner gave. It comes up all the time.
$endgroup$
– Ross Millikan
Jan 21 at 2:10

add a comment | 

3 Answers
3

active

oldest

votes

1

$begingroup$

The answer is $boxed{205}$

Let the two perfect squares be $m$ and $n$. Then, by the statement of the problem, we must have $$m^{2} - n^{2} = (m - n)(m + n)= 133 = 7 cdot 19,$$ where the first equality follows from difference of squares, and the last equality follows from the prime factorization of $133$.

Thus, we must have exactly one of $m - n$ and $m + n$ to equal $7$ and the other to equal $19$. There are two cases to consider:

Case 1: $m + n = 7$ and $m - n = 19$ If this is true, we must have $m = 13$ and $n = -6$, giving us $13^2 + (-6)^2 = 205$

Case 2: $m + n = 19$ and $m - n = 7$. In this case, we have $m = 13$ and $n = 6$, which again gives us a sum of $205$.

Finally, check border cases and note we can get $m^2 - n^2 = 67^2 - 66^2 = 133,$ giving the sum $8845$.

Since $text{min}(205, 205, 8845) = 205$, the answer is $205$.

share|cite|improve this answer

edited Jan 21 at 15:57

answered Jan 21 at 2:58

Ekesh KumarEkesh Kumar

1,04928

$endgroup$

• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  The answer is not $7$. The question asks for min$(m^2+n^2)$, which is $13^2+6^2=205$
  $endgroup$
  – Daniel Mathias
  Jan 21 at 15:16
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  Thanks I fixed it
  $endgroup$
  – Ekesh Kumar
  Jan 21 at 15:42
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  Also note: $67^2-66^2=133$, which gives the higher sum of $8845$
  $endgroup$
  – Daniel Mathias
  Jan 21 at 15:48
  

  
  
  
  

add a comment | 

0

$begingroup$

Hint

The divisors of $133$ are $1, 7,19,133$

Can you use this with the clues from the comments?

share|cite|improve this answer

answered Jan 21 at 2:15

WaveXWaveX

2,7372722

$endgroup$

add a comment | 

0

$begingroup$

To spell it out:

$m^2 - n^2 = 133$

$(m-n)(m+n) = 1*133 = 7*19$.

So .....

share|cite|improve this answer

answered Jan 21 at 2:19

fleabloodfleablood

71.9k22686

$endgroup$

add a comment | 

3 Answers
3

active

oldest

votes

3 Answers
3

active

oldest

votes

active

oldest

votes

active

oldest

votes

1

$begingroup$

The answer is $boxed{205}$

Let the two perfect squares be $m$ and $n$. Then, by the statement of the problem, we must have $$m^{2} - n^{2} = (m - n)(m + n)= 133 = 7 cdot 19,$$ where the first equality follows from difference of squares, and the last equality follows from the prime factorization of $133$.

Thus, we must have exactly one of $m - n$ and $m + n$ to equal $7$ and the other to equal $19$. There are two cases to consider:

Case 1: $m + n = 7$ and $m - n = 19$ If this is true, we must have $m = 13$ and $n = -6$, giving us $13^2 + (-6)^2 = 205$

Case 2: $m + n = 19$ and $m - n = 7$. In this case, we have $m = 13$ and $n = 6$, which again gives us a sum of $205$.

Finally, check border cases and note we can get $m^2 - n^2 = 67^2 - 66^2 = 133,$ giving the sum $8845$.

Since $text{min}(205, 205, 8845) = 205$, the answer is $205$.

share|cite|improve this answer

edited Jan 21 at 15:57

answered Jan 21 at 2:58

Ekesh KumarEkesh Kumar

1,04928

$endgroup$

• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  The answer is not $7$. The question asks for min$(m^2+n^2)$, which is $13^2+6^2=205$
  $endgroup$
  – Daniel Mathias
  Jan 21 at 15:16
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  Thanks I fixed it
  $endgroup$
  – Ekesh Kumar
  Jan 21 at 15:42
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  Also note: $67^2-66^2=133$, which gives the higher sum of $8845$
  $endgroup$
  – Daniel Mathias
  Jan 21 at 15:48
  

  
  
  
  

add a comment | 

1

$begingroup$

The answer is $boxed{205}$

Let the two perfect squares be $m$ and $n$. Then, by the statement of the problem, we must have $$m^{2} - n^{2} = (m - n)(m + n)= 133 = 7 cdot 19,$$ where the first equality follows from difference of squares, and the last equality follows from the prime factorization of $133$.

Thus, we must have exactly one of $m - n$ and $m + n$ to equal $7$ and the other to equal $19$. There are two cases to consider:

Case 1: $m + n = 7$ and $m - n = 19$ If this is true, we must have $m = 13$ and $n = -6$, giving us $13^2 + (-6)^2 = 205$

Case 2: $m + n = 19$ and $m - n = 7$. In this case, we have $m = 13$ and $n = 6$, which again gives us a sum of $205$.

Finally, check border cases and note we can get $m^2 - n^2 = 67^2 - 66^2 = 133,$ giving the sum $8845$.

Since $text{min}(205, 205, 8845) = 205$, the answer is $205$.

share|cite|improve this answer

edited Jan 21 at 15:57

answered Jan 21 at 2:58

Ekesh KumarEkesh Kumar

1,04928

$endgroup$

• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  The answer is not $7$. The question asks for min$(m^2+n^2)$, which is $13^2+6^2=205$
  $endgroup$
  – Daniel Mathias
  Jan 21 at 15:16
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  Thanks I fixed it
  $endgroup$
  – Ekesh Kumar
  Jan 21 at 15:42
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  Also note: $67^2-66^2=133$, which gives the higher sum of $8845$
  $endgroup$
  – Daniel Mathias
  Jan 21 at 15:48
  

  
  
  
  

add a comment | 

1

1

1

$begingroup$

The answer is $boxed{205}$

Let the two perfect squares be $m$ and $n$. Then, by the statement of the problem, we must have $$m^{2} - n^{2} = (m - n)(m + n)= 133 = 7 cdot 19,$$ where the first equality follows from difference of squares, and the last equality follows from the prime factorization of $133$.

Thus, we must have exactly one of $m - n$ and $m + n$ to equal $7$ and the other to equal $19$. There are two cases to consider:

Case 1: $m + n = 7$ and $m - n = 19$ If this is true, we must have $m = 13$ and $n = -6$, giving us $13^2 + (-6)^2 = 205$

Case 2: $m + n = 19$ and $m - n = 7$. In this case, we have $m = 13$ and $n = 6$, which again gives us a sum of $205$.

Finally, check border cases and note we can get $m^2 - n^2 = 67^2 - 66^2 = 133,$ giving the sum $8845$.

Since $text{min}(205, 205, 8845) = 205$, the answer is $205$.

share|cite|improve this answer

edited Jan 21 at 15:57

answered Jan 21 at 2:58

Ekesh KumarEkesh Kumar

1,04928

$endgroup$

The answer is $boxed{205}$

Let the two perfect squares be $m$ and $n$. Then, by the statement of the problem, we must have $$m^{2} - n^{2} = (m - n)(m + n)= 133 = 7 cdot 19,$$ where the first equality follows from difference of squares, and the last equality follows from the prime factorization of $133$.

Thus, we must have exactly one of $m - n$ and $m + n$ to equal $7$ and the other to equal $19$. There are two cases to consider:

Case 1: $m + n = 7$ and $m - n = 19$ If this is true, we must have $m = 13$ and $n = -6$, giving us $13^2 + (-6)^2 = 205$

Case 2: $m + n = 19$ and $m - n = 7$. In this case, we have $m = 13$ and $n = 6$, which again gives us a sum of $205$.

Finally, check border cases and note we can get $m^2 - n^2 = 67^2 - 66^2 = 133,$ giving the sum $8845$.

Since $text{min}(205, 205, 8845) = 205$, the answer is $205$.

share|cite|improve this answer

edited Jan 21 at 15:57

answered Jan 21 at 2:58

Ekesh KumarEkesh Kumar

1,04928

share|cite|improve this answer

share|cite|improve this answer

edited Jan 21 at 15:57

edited Jan 21 at 15:57

edited Jan 21 at 15:57

answered Jan 21 at 2:58

Ekesh KumarEkesh Kumar

1,04928

answered Jan 21 at 2:58

Ekesh KumarEkesh Kumar

1,04928

answered Jan 21 at 2:58

Ekesh KumarEkesh Kumar

1,04928

1,04928

• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  The answer is not $7$. The question asks for min$(m^2+n^2)$, which is $13^2+6^2=205$
  $endgroup$
  – Daniel Mathias
  Jan 21 at 15:16
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  Thanks I fixed it
  $endgroup$
  – Ekesh Kumar
  Jan 21 at 15:42
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  Also note: $67^2-66^2=133$, which gives the higher sum of $8845$
  $endgroup$
  – Daniel Mathias
  Jan 21 at 15:48
  

  
  
  
  

add a comment | 

• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  The answer is not $7$. The question asks for min$(m^2+n^2)$, which is $13^2+6^2=205$
  $endgroup$
  – Daniel Mathias
  Jan 21 at 15:16
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  Thanks I fixed it
  $endgroup$
  – Ekesh Kumar
  Jan 21 at 15:42
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  Also note: $67^2-66^2=133$, which gives the higher sum of $8845$
  $endgroup$
  – Daniel Mathias
  Jan 21 at 15:48
  

  
  
  
  

$begingroup$
The answer is not $7$. The question asks for min$(m^2+n^2)$, which is $13^2+6^2=205$
$endgroup$
– Daniel Mathias
Jan 21 at 15:16

$begingroup$
The answer is not $7$. The question asks for min$(m^2+n^2)$, which is $13^2+6^2=205$
$endgroup$
– Daniel Mathias
Jan 21 at 15:16

$begingroup$
Thanks I fixed it
$endgroup$
– Ekesh Kumar
Jan 21 at 15:42

$begingroup$
Thanks I fixed it
$endgroup$
– Ekesh Kumar
Jan 21 at 15:42

$begingroup$
Also note: $67^2-66^2=133$, which gives the higher sum of $8845$
$endgroup$
– Daniel Mathias
Jan 21 at 15:48

$begingroup$
Also note: $67^2-66^2=133$, which gives the higher sum of $8845$
$endgroup$
– Daniel Mathias
Jan 21 at 15:48

add a comment | 

0

$begingroup$

Hint

The divisors of $133$ are $1, 7,19,133$

Can you use this with the clues from the comments?

share|cite|improve this answer

answered Jan 21 at 2:15

WaveXWaveX

2,7372722

$endgroup$

add a comment | 

0

$begingroup$

Hint

The divisors of $133$ are $1, 7,19,133$

Can you use this with the clues from the comments?

share|cite|improve this answer

answered Jan 21 at 2:15

WaveXWaveX

2,7372722

$endgroup$

add a comment | 

0

0

0

$begingroup$

Hint

The divisors of $133$ are $1, 7,19,133$

Can you use this with the clues from the comments?

share|cite|improve this answer

answered Jan 21 at 2:15

WaveXWaveX

2,7372722

$endgroup$

Hint

The divisors of $133$ are $1, 7,19,133$

Can you use this with the clues from the comments?

share|cite|improve this answer

answered Jan 21 at 2:15

WaveXWaveX

2,7372722

share|cite|improve this answer

share|cite|improve this answer

answered Jan 21 at 2:15

WaveXWaveX

2,7372722

answered Jan 21 at 2:15

WaveXWaveX

2,7372722

answered Jan 21 at 2:15

WaveXWaveX

2,7372722

2,7372722

add a comment | 

add a comment | 

0

$begingroup$

To spell it out:

$m^2 - n^2 = 133$

$(m-n)(m+n) = 1*133 = 7*19$.

So .....

share|cite|improve this answer

answered Jan 21 at 2:19

fleabloodfleablood

71.9k22686

$endgroup$

add a comment | 

0

$begingroup$

To spell it out:

$m^2 - n^2 = 133$

$(m-n)(m+n) = 1*133 = 7*19$.

So .....

share|cite|improve this answer

answered Jan 21 at 2:19

fleabloodfleablood

71.9k22686

$endgroup$

add a comment | 

0

0

0

$begingroup$

To spell it out:

$m^2 - n^2 = 133$

$(m-n)(m+n) = 1*133 = 7*19$.

So .....

share|cite|improve this answer

answered Jan 21 at 2:19

fleabloodfleablood

71.9k22686

$endgroup$

To spell it out:

$m^2 - n^2 = 133$

$(m-n)(m+n) = 1*133 = 7*19$.

So .....

share|cite|improve this answer

answered Jan 21 at 2:19

fleabloodfleablood

71.9k22686

share|cite|improve this answer

share|cite|improve this answer

answered Jan 21 at 2:19

fleabloodfleablood

71.9k22686

answered Jan 21 at 2:19

fleabloodfleablood

71.9k22686

answered Jan 21 at 2:19

fleabloodfleablood

71.9k22686

71.9k22686

add a comment | 

add a comment | 

This page is only for reference, If you need detailed information, please check here