2016/064)Find all solutions in integers $x,y$ of the equation $y^2+2y= x^4+20x^3+104x^2 + 40x + 2003$(Irish 2003 paper 2)

Add 1 to both sides to get
$(y+1)^2 = x^4+20x^3+104x^2+40x+2004 = x^4+20x^3+ 104x^2+40x+ 4 + 2000 $
$= (x^2+10x+2)^2 +2000$
or $(y+1)^2 - (x^2+10x+2)^2 = 2000$
for the above to have solution we need to have $(y+1)$ and $(x^2+10x+2)$ both should be positive and as $(x^2+10x+2) = (x+5)^2-23$
so the lower number need to be 23 less than a perfect square.
so let us find (t,z) which are $(\pm501,\pm499),(\pm252,\pm248),(\pm129,\pm121),(\pm105,\pm95),(\pm60,\pm40),(\pm45,\pm5)$ out of which
only z = 121 which 23 less than is a perfect square
so we get
$y+1= \pm 129, (x+5) = \pm 12$ giving 4 solutions ($y=-130, x = - 17$), ($y= -130,x= 7$), ($y=128,x=7$),($y= 128, x= -17$)

2016/063) Prove for all integers $N>1$ $(N^2)^{2014}- (N^{11})^{106}$ is divisible by $N^6+ N^3+1$

$(N^2)^{2014}- (N^{11})^{106}$
$=N^{4028} - N^{1166}$
$=N^{1166}(N^{2862}-1) = N^{1166}((N^9)^{318}-1)$ is divisible by $N^9-1$
as $N^9-1 = (N^3)^3 -1 = (N^3-1)(N^6+N^3+1)$
as $(N^2)^{2014}- (N^{11})^{106}$ is divsible by $N^9-1$ which is divisible by $N^6+N^3+1$ hence  $(N^2)^{2014}- (N^{11})^{106}$
is divisible by $N^6+N^3+1$

2016/062) If $x= 2 + \sqrt[3]{2^2} + \sqrt[3]{2}$ then the value of $x^3 - 6x^2+6x$ is

we have  $x-  2 =  \sqrt[3]{2^2} + \sqrt[3]{2}$
hence $(x-2)^3 = 4 + 2 + 3 * 2 *(\sqrt[3]{2^2} + \sqrt[3]{2}) = 6 + 3 * 2 (x-2) = 6x - 6$
Hence $x^3 - 6x^2 + 12x -8 = 6x-6 $ or $x^3-6x^2+6x = 2$

2016/061)Find the smallest positive integer m such that 5m is an exact 5th power, 6m is an exact 6th power, and 7m is an exact 7th power. (26th Irish)

The number has to be of the form $5^a6^b7^c$
now $5^{a+1}6^b7^c$ is a 5th power so  $a+1 \equiv 0 \pmod 5$ ,$b \equiv 0 \pmod 5$,$c \equiv 0 \pmod 5$
$5^a6^{b+1}7^c$ is a 6th power so  $a \equiv 0 \pmod 6$ ,$b+1  \equiv 0 \pmod 6$,$c \equiv 0 \pmod 6$
$5^a6^b7^{c+1}$ is a 7th power so  $a \equiv 0 \pmod 7$ ,$b \equiv 7 \pmod 5$,$c+1 \equiv 0 \pmod 7$
so we need to solve for
$a+1 \equiv 0 \pmod 5$ ,$a \equiv 0 \pmod 42$ giving a = 84 (taking multiples of 42 adding 1 to be divsible by 5)
$b+1 \equiv 0 \pmod 6$ ,$b \equiv 0 \pmod 35$ giving b = 35 (taking multiples of 35 adding 1 to be divsible by 6)
$c+1 \equiv 0 \pmod 7$ ,$c \equiv 0 \pmod 30$ giving c = 90 (taking multiples of 30 adding 1 to be divsible by 7)
so the number is $5^{84}* 6^{35}*c^{90}$

2016/060) Find all integers x such that $x(x+1)(x+7)(x+8)$ is square of an integer (21st Irish math olympiad)

we see that x = 0, x = -1, x = -7 and x = -8 gives the answer zero so a perfect square
let us look for other values
we have
$x(x+1)(x+7)(x+8)$
= $x(x+8)(x+1)(x+7)$
= $(x^2+8x)(x^2+8x+7)$
$= y(y+7)$ where y is $x^2+8x$
for it to be a perfect square we see that $(GCD(y,y+7) = GCD(y,7)$
y cannot be a multiple of if 7 beacuse then y and y + 7 are consecutive multiples of 7 and as y is not zero product cannot
be a perfect square.
so y and y + 7 are coprimes and hence perfect square or -ve of perfect square
let $y = n^2$ and $y+7 = m^2$
giving $n^2+7=m^2$
or $m^2-n^2 = 7$
or $(m+n)(n-n) = 7 * 1$ hence $m+n = 7, m-n= 1$ or $m= 4,n= 3$
so $y = 16$
hence $x^2+8x-9=0$ giving $x = 1,=9$
taking -ve values we have $y= - 16 , y + 7 = - 9$
or $x^2+8x+ 16= 0 => x = - 4$
so we have x is one of -9,-8,-7, -1,0,1$

2016/059) Show that in 10 consecutive numbers there is at least one number which is co-prime to other 9 numbers

Out of 10 consecutive numbers 5 numbers are divisible by 2 and not more that 4 numbers are divisible by 3 out of which maximum 2 numbers are odd and divisible by by 3 that makes 7, 2 numbers are divisible by 5 out of which is even so there are maximum one number is divisible by 5 and neither 2 nor 3 and that nakes 8 and maximum 2 numbers are divisible by 7 out of which is only one is odd maximum one number is divisible by 7 and neither 2 nor 3 nor 5 and that makes maximum 9 numbers that are divisible by one of 2,3,5,7. so there is at least one number which is not divisible by 2,3,5 or 7 so the lowest prime factor of the same is 11 and it cannot devide any other number of the set. so this number is co-prime to rest 9.

2016/058) Show that $\frac{1}{n}+\frac{1}{n+1}+\frac{1}{n+2}+\cdots\frac{1}{2n-1}$

=$1-\frac{1}{2} + \frac{1}{3} + \frac{1}{4}\cdots -\frac{1}{2n-1}$
 We have RHS = $\sum_{k=1}^{n}\frac{1}{2k-1} - \sum_{k=1}^{n}\frac{1}{2n}$ $=\sum_{k=1}^{n}\frac{1}{2n-1} + \sum_{k=1}^{n}\frac{1}{2k} - 2 \sum_{k=1}^{n-1}\frac{1}{2k}$ $=\sum_{k=1}^{2n-1}\frac{1}{k}- \sum_{k=1}^{n-1}\frac{1}{k}$ $=\sum_{k=n}^{2n-1}\frac{1}{k}=LHS$