How to prove and find all natural numbers that make $a^2 - b^2 = 5$? Announcing the arrival of...
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How to prove and find all natural numbers that make $a^2 - b^2 = 5$?
Announcing the arrival of Valued Associate #679: Cesar Manara
Planned maintenance scheduled April 23, 2019 at 23:30 UTC (7:30pm US/Eastern)Prove or disprove that $a$ and $b$ are coprime integers iff $a^2$ and $b^2$ are coprime integersProve that if ${a_n}$ is a sequence of natural numbers such that ${ a_n^{-1}}$ is an arithmetic sequence then all $a_i$ are equalFind any natural number $n>1$, such that $n^{100} < 2^n $Find all natural numbers n such that $phi(n)$=12Finding all natural numbers $ $x and $ y$?Find all integers such that ϕ(n) =n/2Infinite probability of natural numbers.Coprime numbers and common divisorsHow to (dis)prove that for all prime numbers $p$, $p$ divides the binomial coefficient $binom pk$How to count all restricted partitions of the number $155$ into a sum of $10$ natural numbers between $[0,30]$?
$begingroup$
Whilst I know that $a = 3$ and $b = 2$, I have no clue how to prove it.
elementary-number-theory
edited Mar 26 at 11:07
P. Schulze
749
asked Mar 26 at 10:54
li lili li
1
$endgroup$
add a comment |
$begingroup$
Whilst I know that $a = 3$ and $b = 2$, I have no clue how to prove it.
elementary-number-theory
edited Mar 26 at 11:07
P. Schulze
749
asked Mar 26 at 10:54
li lili li
1
$endgroup$
10
$begingroup$
Hint: $a^2-b^2=(a-b)(a+b)$.
$endgroup$
– lulu
Mar 26 at 10:56
1
$begingroup$
Hint 2: 5 is prime so any product a*b = 5 has to be a=1 and b=5
$endgroup$
– P. Schulze
Mar 26 at 10:59
$begingroup$
In a more general setting where nonpositive integers are also allowed, you get four different solutions instead, $(a,b)in { (-3, -2), (-3, 2), (3, -2), (3, 2) }$. Can you see why?
$endgroup$
– Jeppe Stig Nielsen
Mar 26 at 11:04
$begingroup$
Another tack: The difference between two consecutive squares is $(n+1)^2 - n^2 = 2n+1.$ So you must have $2n+1 leq 5.$ That leaves you with a very finite set of cases.
$endgroup$
– B. Goddard
Mar 27 at 13:32
add a comment |
$begingroup$
Whilst I know that $a = 3$ and $b = 2$, I have no clue how to prove it.
elementary-number-theory
edited Mar 26 at 11:07
P. Schulze
749
asked Mar 26 at 10:54
li lili li
1
$endgroup$
Whilst I know that $a = 3$ and $b = 2$, I have no clue how to prove it.
elementary-number-theory
elementary-number-theory
edited Mar 26 at 11:07
P. Schulze
749
asked Mar 26 at 10:54
li lili li
1
edited Mar 26 at 11:07
P. Schulze
749
asked Mar 26 at 10:54
li lili li
1
edited Mar 26 at 11:07
P. Schulze
749
edited Mar 26 at 11:07
P. Schulze
749
edited Mar 26 at 11:07
P. Schulze
749
749
asked Mar 26 at 10:54
li lili li
1
asked Mar 26 at 10:54
li lili li
1
asked Mar 26 at 10:54
li lili li
1
1
10
$begingroup$
Hint: $a^2-b^2=(a-b)(a+b)$.
$endgroup$
– lulu
Mar 26 at 10:56
1
$begingroup$
Hint 2: 5 is prime so any product a*b = 5 has to be a=1 and b=5
$endgroup$
– P. Schulze
Mar 26 at 10:59
$begingroup$
In a more general setting where nonpositive integers are also allowed, you get four different solutions instead, $(a,b)in { (-3, -2), (-3, 2), (3, -2), (3, 2) }$. Can you see why?
$endgroup$
– Jeppe Stig Nielsen
Mar 26 at 11:04
$begingroup$
Another tack: The difference between two consecutive squares is $(n+1)^2 - n^2 = 2n+1.$ So you must have $2n+1 leq 5.$ That leaves you with a very finite set of cases.
$endgroup$
– B. Goddard
Mar 27 at 13:32
add a comment |
10
$begingroup$
Hint: $a^2-b^2=(a-b)(a+b)$.
$endgroup$
– lulu
Mar 26 at 10:56
1
$begingroup$
Hint 2: 5 is prime so any product a*b = 5 has to be a=1 and b=5
$endgroup$
– P. Schulze
Mar 26 at 10:59
$begingroup$
In a more general setting where nonpositive integers are also allowed, you get four different solutions instead, $(a,b)in { (-3, -2), (-3, 2), (3, -2), (3, 2) }$. Can you see why?
$endgroup$
– Jeppe Stig Nielsen
Mar 26 at 11:04
$begingroup$
Another tack: The difference between two consecutive squares is $(n+1)^2 - n^2 = 2n+1.$ So you must have $2n+1 leq 5.$ That leaves you with a very finite set of cases.
$endgroup$
– B. Goddard
Mar 27 at 13:32
10
10
$begingroup$
Hint: $a^2-b^2=(a-b)(a+b)$.
$endgroup$
– lulu
Mar 26 at 10:56
$begingroup$
Hint: $a^2-b^2=(a-b)(a+b)$.
$endgroup$
– lulu
Mar 26 at 10:56
1
1
$begingroup$
Hint 2: 5 is prime so any product a*b = 5 has to be a=1 and b=5
$endgroup$
– P. Schulze
Mar 26 at 10:59
$begingroup$
Hint 2: 5 is prime so any product a*b = 5 has to be a=1 and b=5
$endgroup$
– P. Schulze
Mar 26 at 10:59
$begingroup$
In a more general setting where nonpositive integers are also allowed, you get four different solutions instead, $(a,b)in { (-3, -2), (-3, 2), (3, -2), (3, 2) }$. Can you see why?
$endgroup$
– Jeppe Stig Nielsen
Mar 26 at 11:04
$begingroup$
In a more general setting where nonpositive integers are also allowed, you get four different solutions instead, $(a,b)in { (-3, -2), (-3, 2), (3, -2), (3, 2) }$. Can you see why?
$endgroup$
– Jeppe Stig Nielsen
Mar 26 at 11:04
$begingroup$
Another tack: The difference between two consecutive squares is $(n+1)^2 - n^2 = 2n+1.$ So you must have $2n+1 leq 5.$ That leaves you with a very finite set of cases.
$endgroup$
– B. Goddard
Mar 27 at 13:32
$begingroup$
Another tack: The difference between two consecutive squares is $(n+1)^2 - n^2 = 2n+1.$ So you must have $2n+1 leq 5.$ That leaves you with a very finite set of cases.
$endgroup$
– B. Goddard
Mar 27 at 13:32
add a comment |
0
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10
$begingroup$
Hint: $a^2-b^2=(a-b)(a+b)$.
$endgroup$
– lulu
Mar 26 at 10:56
1
$begingroup$
Hint 2: 5 is prime so any product a*b = 5 has to be a=1 and b=5
$endgroup$
– P. Schulze
Mar 26 at 10:59
$begingroup$
In a more general setting where nonpositive integers are also allowed, you get four different solutions instead, $(a,b)in { (-3, -2), (-3, 2), (3, -2), (3, 2) }$. Can you see why?
$endgroup$
– Jeppe Stig Nielsen
Mar 26 at 11:04
$begingroup$
Another tack: The difference between two consecutive squares is $(n+1)^2 - n^2 = 2n+1.$ So you must have $2n+1 leq 5.$ That leaves you with a very finite set of cases.
$endgroup$
– B. Goddard
Mar 27 at 13:32