2021/038) if $n^2+7n+4$ is divisible by 11 what is the remainder when divided by 121

 Working in mod 11 we have

$n^2+7n + 4 \equiv n^2 + (7-11)n + 4$

$\equiv n^2 - 4n + 4 \equiv (n-2)^2$

so if $n^2 + 7n + 4$ is divisible by 11 then n-2 is divisible by 11

so $n = 11k + 2$

So  $n^2+7n+4 = (11k+2)^2 + 7(11k+2) + 4$

= $121k^2 + 121k + 22$

so remainder when divided  by 121 is 22 

2021/037) Solve the system of equations

$a + b + c = 7\cdots(1)$

$a^2 + b^2 + c^2 = 21\cdots(2)$

$abc= 8 \cdots(3)$

Solution 

We are given 

$a + b + c = 7\cdots(1)$

$a^2 + b^2 + c^2 = 21\cdots(2)$

$abc= 8 \cdots(3)$

From (1) and (2) we get

$(a+b+c)^2 - (a^2+b^2+ c^2) = 7^2 - 21$

Or $2(ab+bc+ca) = 28$

Or $ab+bc+ca = 14\cdots(4)$

a , b, c are roots of equation

$(x-a)(x-b)(x-c) = 0$

Or $x^3-x^2(a+b+c) + x(ab + bc + ca) - abc= 0$

Putting the values from (1), (3) and (4) we get 

$x^3 - 7x^2 + 14x - 8 = 0$

We know from above that the value becomes zero when x = 1 so x =1 is a factor and (dividing by x- 1) we get

$(x-1)(x^2 - 6x + 8) = 0$

Or $(x-1)(x-2)(x-4) = 0$

So roots are 1,2, 4 and so we have

$\{a,b,c\} = \{1,2,4\}$ 

2021/036) Simplify : $\sqrt{4 -4a + a^2} + \sqrt{4 + 4a + a^2}$, if $a < -2 $

We have $\sqrt{4 -4a + a^2} = \sqrt{(a-2)^2}= | a- 2 | $

as $a < -2 $ so  $\sqrt{4 -4a + a^2} = 2 - a$

Further we have $\sqrt{4 + 4a + a^2} = \sqrt{(a+2)^2}= | a + 2 | $

as $a < -2 $ so  $\sqrt{4 -4a + a^2} = - 2 - a$

adding we get $\sqrt{4 -4a + a^2}  +  \sqrt{4 + 4a + a^2}= - 2a$ 

2021/035) Let $x,\,y,\,z$ be integers such that $(x-y)^2+(y-z)^2+(z-x)^2=xyz$, prove that $x^3+y^3+z^3$ is divisible by $x+y+z+6$.

 We know $x^3+y^3+z^3 - 3xyz = \frac{1}{2}(x+y+z)((x-y)^2 + (y-z)^2 + (z-x)^2)$

Hence  $x^3+y^3+z^3 = 3xyz + \frac{1}{2}(x+y+z)((x-y)^2 + (y-z)^2 + (z-x)^2)$

Hence $x^3+y^3+z^3 = 3xyz + \frac{1}{2}(x+y+z)(xyz)$ (putting the value from given condition)

Or $x^3+y^3+z^3 = xyz( 3 + \frac{1}{2}(x+y+z))$

Or $x^3+y^3+z^3 = \frac{xyz}{2}( 6 + x+y+z)$

If we can prove that xyz is even then  we are through

As (x-y), (y-z) and (z-x) sum to give zero so atleast one of them is even. So xyz is even from the given condition so  $\frac{xyz}{2}$ is an integer and hence $x^3+y^3+z^3$ is multiple of $(6 + x+y+z)$

2021/034)How do you prove if a, b, and c are integers such that $a|b$ and $c|b$ and a and c are co-primes , then $ac | b$

 Now  $a|b$  so there exists integer d such that $$b = ad\cdots(1)$$

Further  c and a are co-primes so as ber Bezout's identity

There exists integer x and y such that

$cx + ay = 1$

Multiply both sides by d to get

$cx + ayd = d$

or $cx + by = d$ using (1)

as c divides b so there exists integer p such that

$b= pc$

So $cx + pcy = d$

Or $c(x+py) = d$

So from (1) $b = ad = ac(x+py)$ hence ac is factor of b

2021/033) Solve the system of equations

$xy = z - x - y\cdots(1)$

$yz = x - y - z\cdots(2)$

$zx = y - x - z\cdots(3)$

From (1) we have

Solution

x + y + xy = z

or xy + x + y + 1 = z + 1

or $(x+1)(y+1) = z+1\cdots(4)$

similarly from (2) and (3) get

 $(y+1)(z+1) = x+1\cdots(5)$

 $(z+1)(x+1) = y+1\cdots(6)$

 multiplying (4),(5),(56) we get

 $(x+1)^2(y+1)^2(z+1)^2 = (x+1)(y+1)(z+1)$

 so $(x+1)(y+1)(z+1) = 1$ or $(x+1)(y+1)(z+1) = 0$

  $(x+1)(y+1)(z+1) = 0$ gives x+ 1 = 0 or y + 1 = 0 or z + 1 = 0

  x+1 = 0 using (4) and (6)  y+ 1 = z+1 =0

 Similarly y + 1 = 0 gives  x+ 1 = z+1 =0

    and  z + 1 = 0 gives x+1 =  + 1 =0 

    so solution x = -1, y  = -1, z = -1

  product 1 along with (1), (2), (3) give x+1 = y+ 1 = z + 1 = 1 ( that is x= 1, y = 1, z = 1)

  or x + 1 = y+ 1 = -1 and z + 1 = 1( x=y=-2,z = 0)

 or y + 1 = z+ 1 = -1 and x + 1 = 1( y=z = -2,x = 0)

 or x + 1 = z+ 1 = -1 and y + 1 = 1( x=z=-2,y = 0) 

2021/032) Find non -ve integer x and y satisfying $x^3 = y^3 + 2y + 1$

We have $(y+2)^3 = y^3 + 6y^2+12y + 8 > y^3 + 2y + 1$

and $y^3 < y^3 + 2y + 1$

so x = y + 1

or $(y+1)^3 = y^2 + 3y^2 + 3y + 1 = y^2 + 2y + 1$

or  $2y^2 + y = 0$ giving y = 0(other root -ve and so not admissible)  and x = 1

2021/031) Given $(4-a)(4-b)(4-c)(4-d)(4-e)=12$ a,b,c,d,e are negative or positive integers which have different values Find the sum of a+b+c+d+e

First let us factor 12 we have 12 = 2 * 2 * 3.

The above has 3 factors and given expression has 5 factors so
12 = 1 * 1 * 2 * 2 * 3
In the above we ave 5 factors but not all 5 are different and to make them different we can have (knowing that -ve numbers are allowed
12 = 1 * (-1) * 2 * (-2) * 3
The (4-a,4-b,4-c,4-d,4-e) can be chosen as permutation of(1,-1,2,-2,3) giving a sum 3 and so a+b+c+d+e = 20 -3 or 17.

2021/030)Find prime p and positive integers x,y such that $p^x = y^4 + 4$

We get factoring RHS 

$p^x = y^4 + 4 = (y^4 + 4y^2 + 4) - 4y^2$

$= (y^2 - 2y +2)(y^2+ 2y2)$

as RHS is a power of p so each factor is power of p

and clearly $ (y^2 - 2y +2) < (y^2+ 2y+2)$

so $(y^2 - 2y +2) |  (y^2+ 2y+2)\cdots(1)$

Or $(y^2 - 2y +2) |  (y^2 - 2y +2) -  (y^2+ 2y+2)$

Or $(y^2 - 2y +2) |  (y^2 - 2y+2) -  (y^2+ 2y+2)$

Or $(y^2- 2y+2 | 4y\cdots(2)$

Hence $(y^2- 2y+2 | 4y^2\cdots(3)$

From (1) and (3) we get

$(y^2 - 2y +2) |  4(y^2+ 2y+2)- 4y^2$

or $(y^2 - 2y +2) |  8y+8 $

$(y^2 - 2y +2) |  8(y + 1)\cdots(4) $

As y and y+ 1 are co-primes from 2) ad (4)

$(y^2 - 2y +2) |  8\cdots(4) $

 

So we have 4 cases $y^2-2y+2$ = 1 or 2 or 4 or 8

Let us take the csses one by 1

case 1)

$y^2-2y+2 = 1$

or $y^2-2y+1= 0$ or $y= 1$

this gives $p^x = 1^4 + 4 = 5$ giving $p=5,x = 1$

 Case 2)

 

$y^2-2y+2 = 2$

or $y^2-2y = 0$ or $y= 0$ or 2 

$y=2$ gives $p^x = 20$ which s not a power of a prime

Case 3)

 

$y^2-2y+2 = 4$

or $y^2 - 2y - 2= 0$ this does not have integer solution

Case 4)

 

$y^2-2y + 2 = 8$

or $y^2 - 2y - 6= 0$ and this does not have integer solution

So the solution set $p=5,x=1,y=1$

 

2021/029) If (a,b,c) is a Pythagorean triple prove that $60 | abc$

 we have 60 = 3 * 4 * 5.

For 60 to be factor of abc we need to show that 3 is a factor of a or b or c. 4 is a factor
of a or b or c and 5 is a factor of a or b or c. We shall prove the same for
primitive Pythagorean triple

We now prove the 3 one by one and we shall be proving for primitive triples
1) For primitive Pythagorean triple one of the legs is divisible by 3
Proof:
let no leg be divisible by 3
then we have the 2 sides a and b as $a=3n\pm 1$ and $b=3m\pm 1$
so we have $c^2 = a^2 + b^2= (3n\pm 1)^2 + (3m\pm 1)^2 = (9n^2 \pm 6n + 1) + (9m^2 \pm 6m + 1)$
or $c^2 = 3(3n^2+3m^2 \pm 2n \pm 2m) + 2$
but dividing a square number by 3 does not leave a remainder 2 so this is not possibles. So atleast one side shall be divisible by 3.
2) Exactly one of a, b is divisible by 4
before proving this 1st we need to prove that exactly one of a, b is divisible by 2
because it is primitive so both cannot be divisible by 2 so let us assume that both are odd
so let a = 2x + 1 and b= 2y + 1
so $a^2+b^2 = (2x+1)^2 + (2y+1)^2 = 4x^2 + 4x + 1 + 4y^2 + 4y + 1 = 4(x^2+x+y^2+y) +2$
but this cannot be a perfect square as if it is even then it leaves a remainder 0 when divided by 4.
without loss oe generality let a be odd and b even.
Now a is odd so $a= 2x + 1$
There are 2 cases for b. either b is divisible by 4 or not 5
If b is not divisible by 4 then b is of the form $4y+2$
we have 60 = 3 * 4 * 5.
for 60 to be factor of abc we need to show that 3 is a factor of a or b or c. 4 is a factor of a or b or c and 5 is a factor of a or b or c. We shall prove the same for primitive Pythagorean triple
we now prove the 3 one by one and we shall be proving for primitive triples
1) For primitive Pythagorean triple one of the legs is divisible by 3
Proof:
let no leg be divisible by 3
then we have the 2 sides a and b as $a=3n\pm 1$ and $b=3m\pm 1$
so we have $c^2 = a^2 + b^2= (3n\pm 1)^2 + (3m\pm 1)^2 = (9n^2 \pm 6n + 1) + (9m^2 \pm 6m + 1)$
or $c^2 = 3(3n^2+3m^2 \pm 2n \pm 2m) + 2$
but dividing a square number by 3 does not leave a remainder 2 so this is not possibles. So atleast one side shall be divisible by 3.
2) Exactly one of a, b is divisible by 4
before proving this 1st we need to prove that exactly one of a, b is divisible by 2
because it is primitive so both cannot be divisible by 2 so let us assume that both are odd
so let a = 2x + 1 and b= 2y + 1
so $a^2+b^2 = (2x+1)^2 + (2y+1)^2 = 4x^2 + 4x + 1 + 4y^2 + 4y + 1 = 4(x^2+x+y^2+y) +2$
but this cannot be a perfect square as if it is even then it leaves a remainder 0 when divided by 4.
we have 60 = 3 * 4 * 5.
For 60 to be factor of abc we need to show that 3 is a factor of a or b or c. 4 is a factor
of a or b or c and 5 is a factor of a or b or c. We shall prove the same for
primitive Pythagorean triple
We now prove the 3 one by one and we shall be proving for primitive triples
1) For primitive Pythagorean triple one of the legs is divisible by 3
Proof:
let no leg be divisible by 3
then we have the 2 sides a and b as $a=3n\pm 1$ and $b=3m\pm 1$
so we have $c^2 = a^2 + b^2= (3n\pm 1)^2 + (3m\pm 1)^2 = (9n^2 \pm 6n + 1) + (9m^2 \pm 6m + 1)$
or $c^2 = 3(3n^2+3m^2 \pm 2n \pm 2m) + 2$
but dividing a square number by 3 does not leave a remainder 2 so this is not possibles. So atleast one side shall be divisible by 3.
2) Exactly one of a, b is divisible by 4
before proving this 1st we need to prove that exactly one of a, b is divisible by 2
because it is primitive so both cannot be divisible by 2 so let us assume that both are odd
so let a = 2x + 1 and b= 2y + 1
so $a^2+b^2 = (2x+1)^2 + (2y+1)^2 = 4x^2 + 4x + 1 + 4y^2 + 4y + 1 = 4(x^2+x+y^2+y) +2$
but this cannot be a perfect square as if it is even then it leaves a remainder 0 when divided by 4.
without loss oe generality let a be odd and b even.
Now a is odd so $a= 2x + 1$
There are 2 cases for b. either b is divisible by 4 or not 5
If b is not divisible by 4 then b is of the form $4y+2$
we have 60 = 3 * 4 * 5.
for 60 to be factor of abc we need to show that 3 is a factor of a or b or c. 4 is a factor of a or b or c and 5 is a factor of a or b or c. We shall prove the same for primitive Pythagorean triple
we now prove the 3 one by one and we shall be proving for primitive triples
1) For primitive Pythagorean triple one of the legs is divisible by 3
Proof:
let no leg be divisible by 3
then we have the 2 sides a and b as $a=3n\pm 1$ and $b=3m\pm 1$
so we have $c^2 = a^2 + b^2= (3n\pm 1)^2 + (3m\pm 1)^2 = (9n^2 \pm 6n + 1) + (9m^2 \pm 6m + 1)$
or $c^2 = 3(3n^2+3m^2 \pm 2n \pm 2m) + 2$
but dividing a square number by 3 does not leave a remainder 2 so this is not possibles. So atleast one side shall be divisible by 3.
2) Exactly one of a, b is divisible by 4
before proving this 1st we need to prove that exactly one of a, b is divisible by 2
because it is primitive so both cannot be divisible by 2 so let us assume that both are odd
so let a = 2x + 1 and b= 2y + 1
so $a^2+b^2 = (2x+1)^2 + (2y+1)^2 = 4x^2 + 4x + 1 + 4y^2 + 4y + 1 = 4(x^2+x+y^2+y) +2$
but this cannot be a perfect square as if it is even then it leaves a remainder 0 when divided by 4.
without loss oe generality let a be odd and b even.
Now a is odd so $a= 2x + 1$
There are 2 cases for b. either b is divisible by 4 or not 5
If b is not divisible by 4 then b is of the form $4y+2$
So $a^2 + b^2 = (2x+1)^2 + (4y+2)^2 = 4x^2+4x + 1 + 16y^2 + 16y + 4$
$= 4x(x+1) + 1 + 16y(y+1) + 4$
We have $a^2 + b^2 \equiv 5 \pmod 8$
But because if c is odd say c = 2t + 1
$c^2 = (2t+1) ^2$
Or $c^2 = 4t^2 + 4t + 1 = 4t(t+1) + 1$
So $c^2 \equiv 1 \pmod 8$
Hence $a^2 + b^2 \equiv 5 \pmod 8$ is not possible so b is divisible by 4
3) Either a or b or c is divisible by 5
If either a or b is divisible by 5 then we are through. So let us assume that neither a nor b is divisible by 5 . working in mod 5 we have
$x^2 \in \{0,1,4\} \pmod 5$
as a and b are not divisible by 5 so
$a^2 \in \{1,4\} \pmod 5$
$b^2 \in \{1,4\} \pmod 5$
if we choose $a^2=1 \pmod 5$ and $b^2=4 \pmod 5$ or $a^2=4 \pmod 5$ and $b^2=1 \pmod 5$ then only the sum could be a perfect square and and other combination does not satisfy the criteria working on mod 5 (we get 2 or 3) and this gives $c^2=0 \pmod 5$ or c is divisible by 5.
As we have proved that 3 divides a or b so 3 divides abc, 4 divides a or b so 4 divides abc and 5 divides a or b or c so 5 divides abc. hence 60 is a factor or abc.
Proved

2021/028) Find non -ve integer x and y satisfying $x^3 = y^3 + 2y + 1$

We have $y^3 + 2y + 1 > y^3$

Further $(y+2)^3 = y^3 + 6y^2 + 12y + 8 > y^3 + 2y + 1$ 

So we have $(y+2)^3 > x^3 > y^3$

So $x^3 = (y+1)^3 = y^3 + 3y^2 + 3y +1 = y^3 + 2y + 1$

Or $3y^2 + 3y + 1 = 2y + 1$

Or $3y^2 + y = 0$

Or y = 0 as we are considering non -ve roots

So solution x=1, y = 0