2019/012) if a,b,c are sides of a triangle with the the property that $a^n,b^n,c^n$ are sides of a triangle for each positive integer n then prove that the triangle is isosceles

Without loss of generality let us assume that $a > b>c$

So $a^n < b^n + c^n$
Or $1 < (\frac{b}{a})^n + (\frac{c}{a})^n$
If $a>b > c$ then $(\frac{b}{a})^n = 0$ and $(\frac{c}{a})^n =0$ when n goes to infinite
So $1 < 0$ which is contradiction
So b has to be same as a for the sum to be 1 in the limit
Or the triangle is isosceles

Proved

2019/011) Given a number that does not end in 0 or 5, prove that it has a multiple that consists of only 6's and 7's.

This question I have solved at https://mathhelpboards.com/challenge-questions-puzzles-28/multiple-consists-only-6s-7s-26532.html#post116812

I am providing a structured solution for the same. (not same as in the link above)

There are 2 cases:
1) The number is odd
2) The number is even

 Now we provide solution for both cases to make the solution complete

1) The number is odd say k

The number is not ending with 5 so the number is not divisible by 5

We take numbers $a_1=1, a_2=11, a_3= 111$  so on upto a number with k 1's that is $a_k = 1111....1111$ (k times) that is $a_n$ is a number with n 1's.
Divide the k numbers by k
We have k remainder 0 and so either one remainder is zero or 2 remainder are same (as there are maximum k remainders 0 to k-1)
We assert that one of the remainders is zero.

If that is not the case then 2 remainders are same say for $a_n$ and $a_m$ with $n > m $ so $a_n-a_m = a_{n-m} * 10^m$ divided by k is zero so $a_{n-m}$ divided by k is zero which is a contradiction

so one of the remainders is zero let it be $a_m$ . which is divisible by k. multiply the number bu 6 or 7 so that all the digits are 6 or 7 and it is a multiple of k which is true

case 2: The number is even that is $k*2^n$ Where k  is odd and m is $>0$

Before that we prove the following
We have $10^n \equiv 2^n \pmod {2^{n+1}}\cdots(1)$

Proof:

$10^n = 5^n * 2^n =  \frac{5^n-1}{2} *2^{n+1} + 2^n\equiv  = 2^n \pmod {2^{n+1}}   $ as $\frac{5^n-1}{2}$ is integer

Now 6 is divisible by 2 and 76 by ${2^2}$

If a number is divisible by $2^n$ then either it divisible by $2^{n+1}$ or remainder when divided by $2^{n+1}$ is $2^n$

If it is divisible by $2^{n+1}$ then  add $6*10^{n+1}$ to it to make is divisible by$ 2^{n+1}$ as both terms are divisible by $2^{n+1}$

If it is not divisible by $2^{n+1}$ that is remainder is $2^n$ then  add $7*10^{n+1}$ to it to make is divisible by$ 2^{n+1}$ as both terms leave a remainder $2^n$ and sum zero  divisible by $2^{n+1}$

So in both cases we have a number consisting  of 7 and 6 (n digits) which is divisible by $2^n$

Note: there my be a smaller number satisfying the criteria with m digits (say y) may be $m < n$

So let is consider the number y with p digits ( digits are 6 or 7) which is divisible by $2^n$

Now we choose k numbers $x_1=1 $, $x_2 = 10^p + 1$,$x_3= 10^{2p}+10^p + 1$, $x_4=10^{3p} + 10^{2p} + 10^p + 1 $  so on $x_{k} = 10^{(k-1)p} + 10^{(k-2)p} \cdots + 1 $

By using same argument as in case 1 we can show that

So one of the $x$ above is divisible by k. multiplying by  y we get a number consisting of 7 and 6 which is a multiple of given number

Proved




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2019/010) Given A, B, C, integers satisfying $A \log\, 16 + B \log\, 18 + C \log\, 24 = 0$ find minimum value of $A^2 + B^2+C^2$,

We are given
$A \log\, 16 + B \log\, 18 + C \log\, 24 = 0$
or
$A \log\, 2^4 + B \log\, 2* 3^2 + C \log\, 2^3*3 = 0$
or
$A (4\log\, 2) + B (\log\, 2 + 2\log\,  3) + C (3 \log\, 2 + \log\, 3) = 0$
or $\log\, 2( 4 A+ B + 3C) + \log\, 3(2B + C = 0$
as $\log\,2$ and $\log\,3$ are irrational so we must have

$4A + B + 3C = 0\cdots(1)$
$2B + C = 0\cdots(2)$
From (2)
$C = - 2B\cdots(3)$
And putting in (1) we get $4A= 5B\cdots(4)$

Lowest A, B , - C(all positive)  shall give lowest $A^2+B^2+C^2$
From (4) we get
$A=5, B= 4$ hence $C = -8$ from (3)

So lowest $A^2+B^2+C^2= 25 + 16 + 64 = 105$
(A= -5, B = -4, C= 8) gives same value

2019/009) Find all non-negative integers x,y satisfying $(xy-7)^2 = x^2 + y^2$

We have adding 2xy on both sides

$(xy-7)^2 + 2xy = x^2 +y^2 + 2xy = (x+y)^2$
or $(x^2y^2 -14xy + 49) + 2xy = (x+y)^2$
or $(x^2y^2 -12xy + 49)  = (x+y)^2$
$(x^2y^2 -12xy +36 ) + 13 = (x+y)^2$
or $(xy-6)^2 + 13 = (x+y)^2$
or $13 = (x+y)^2 - (xy-6)^2 = (x+y+xy - 6)(x + y -xy + 6)$
or $13 = ((x+1)(y+1) -7)(5 - (x-1)(y-1))$
this is product of 1 and 13 so we have
$(x+1)(y+1) -7 = 13$ and $5 - (x-1)(y-1) = 1$ giving $x+1)(y+1) = 20$ and $(x-1)(y-1) = 6$
giving (x,y) = (3,4) or (4,3)

Or 
$(x+1)(y+1) -7 = 1$ and $5 - (x-1)(y-1) = 13$ giving $x+1)(y+1) = 8$ and $(x-1)(y-1) = -8$
giving (x,y) = (0,7) or (7,0)

So solution set $(x,y) = (0,7)\, or \,(7,0)\, or\, (3,4)\, or\, (4,3)$

2019/008) Show that $n^2(n^2-1)(n^2-4)$ is divisible by 360

Proof:
We have $n^2(n^2-1)(n^2-4)= n^2(n-1)(n+1)(n-2)(n+2) = (n-2)(n-1)n^2(n+1)(n+2)$
$=5! * {n\choose 5}$

Now $5!=120$ Additionally one of (n-2),(n-1),n is divisible by 3.
If n is divisible by 3 then $n^2$ is divisible by 9 or 2 number (n-2) and (n+1) or (n-1) and (n+2) are divisible by 3.
So the product by $3^2$ so the number is divisible by $LCM(120,9)$ or 360
Or alternatively there is an additional 3 so the product is divisible by 120 *3 or 320