Two Possibilities

Writing $d = \tfrac{ab}{c}$ we see that $a+b+c+d \; = \; a + b + c + \tfrac{ab}{c} \; = \; \frac{(a+c)(b+c)}{c}$ Thus we must be able to write $c = uv$ where $u,v$ are integers such that $u$ divides $a+c$ and $v$ divides $b+c$ . But $a+c > c \ge u$ and $b+c > c \ge v$ and so the integers $m = \tfrac{a+c}{u}$ and $n = \tfrac{b+c}{v}$ are both greater than $1$ , with $a+b+c+d = mn$ . Thus $a+b+c+d$ is not prime.

This is a version of the proof Sathvik Acharya mentions, which I find to be the easiest to visualize. Our general strategy will be to have an area with one side of $a + b + c + d$ that we can rearrange into a new area, which we can use to form a contradiction when assuming $a + b + c + d$ is prime.

First, let's visually represent $a + b + c + d ,$ but slightly out of order as $a + c + d + b$ (just to make a later step easier to visualize).

We want to turn this into an area that we can get factors from. Any of the four integers in the sum work, let's go with $a:$

Now, we're going to make a 2 by 2 box. We can turn and move the $ad$ side to fit into the bottom left. $ab$ is equivalent to $cd$ so we can write that area as the lower right corner.

So we've proven that $a(a+b+c+d) = (a+c)(a+d) .$ This also means the integer $a$ is $\frac{(a+c)(a+d)}{a+b+c+d} .$

Suppose $a + b + c + d$ is prime. Since $a$ is an integer, $a+b+c+d$ must divide evenly into either $(a+c)$ or $(a+d) ;$ however, $a + b + c + d$ is larger than both numbers, so this is a contradiction. Hence $a + b + c + d$ must not be prime.

This solution is based off of @Ram Mohith 's idea, taken to completion.

Let $r$ be a rational number such that $a = rc.$ Since $ab = cd,$ we get $d = rb.$ Now, we find $a + b + c + d$ by substituting in these values for $a, d$ :

$\begin{aligned} a + b + c + d &= rc + b + c + rb \\ &= (b + c)(1 + r). \end{aligned}$

Write $r = \dfrac{r_1}{r_2},$ where $r_1, r_2$ are positive integers such that $\gcd(r_1, r_2) = 1.$ (Note that if $r$ is an integer, $r_2 = 1.$ ) Substituting in this for $r$ yields

$\begin{aligned} a + b + c + d &= (b + c) \left (1 + \dfrac{r_1}{r_2} \right ) \\ &= \dfrac{b + c}{r_2}(r_1 + r_2). \end{aligned}$

We must have $r_2|b, c,$ since $rc$ and $rb$ are integers. Thus, $r_2|b + c,$ and furthermore, $r_2 < b + c$ due to the divisibility, so $\dfrac{b + c}{r_2} > 1.$ This means that $a + b + c + d$ is always the product of two integers greater than 1, so it cannot be prime. $\blacksquare$

For a+b+c+d=p to be prime, p should be odd as it can't be 2(we can verify all cases).

For p to be odd either one or three of a,b,c,d should be odd and others to be even. And ab=cd cannot be satisfied with that, as one side will be odd and other will be even.

So p can't be prime.

It is given that ab = cd

So , $\frac{a}{c}$ = $\frac{d}{b}$

Let , $\frac{a}{c}$ = $\frac{d}{b}$ = k (here k is some constant )

Then , a = ck and d = bk

Now , we have to find the value of a + b+ c + d . So now we will substitute the value of a and c in it .

a + b + c + d = ck + b + c + bk

b + bk + c +ck = b(1 + k) + c(1 + k)

(b + c)(1 + k)

So, we come to know that (a + b + c + d) can be written as product of two numbers (b + c) and (1 + k) out of these two at least one will be an integer .

Therefore we conclude a + b + c + d is not a prime number as it can be written as product of two numbers out of which at least one will be an integer .

Lemma: If $x,y$ are positive integers such that $m \mid xy$ and $m>x,y$ , then $m$ is composite.

Proof: Assume to the contrary that $m$ is prime. Clearly $m$ cannot divide $x$ or $y$ and the condition $m \mid xy$ is not satisfied. This contradiction leads us to the conclusion that $m$ is prime.

Solution: Observe that $a(a+b+c+d)=a^2+ab+ac+ad=a^2+cd+ac+ad=(a+c)(a+d).$ Hence, $a(a+b+c+d)=(a+c)(a+d) \implies a+b+c+d \mid (a+c)(a+d).$ Since $a+b+c+d>((a+c), (a+d))$ from the above lemma we get $a+b+c+d$ is always composite

if ab = cd then if a + b + c + d should be a even number which is not prime number because it is like having 1 + 4 = 2 + 3 and adding two equal numbers will add up to a even number

M=a+b+c+d>=4, if M is a prime number, M need to be an odd number So there two combinations for a, b, c, d. 1)three odd number and one even number, 2) one odd number and three even number. But! How could such two combinations satisfy the equation ab=cd?

For that, we'll need an odd amount of odd numbers, which is impossible if we still want to get ab = cd

if $ab = cd$ , there always exists integers $x,y,z,t$ , such that $a=xy, b=zt, c=xz, d=yt$ so $a+b+c+d = xy+zt+xz+yt = (x+t)(y+z)$ which cannot be prime. we can prove that $x,y,z,t$ exist as follows: $x=\gcd(a,c)$ , $y=\frac{a}{\gcd(a,c)}$ , $z=\frac{c}{\gcd(a,c)}$ , $t=\frac{d\cdot\gcd(a,c)}{a}$ . It is easy to verify that these are all integers and they satisfy above equalities.

ab=cd means if a&b both are odd then c&d both must be odd i. e the sum is a even number and therefore the given sum is not a prime number.

if a&b both are even then c&d both must be even i. e the sum is a even number and therefore the given sum is not a prime number.

How can 3 be equal to 5. It can be proved mathematically though. But the question is how and why?

Let $p=a+b+c+d$ be prime. WLOG for ab and cd, we can write:

$(a+b)^2 = a^2 + 2ab + b^2$

$2ab = (a+b)^2 - a^2-b^2$

We now square the expression a+b+c+d:

$(a+b+c+d)^2$

$a^2 + b^2 + c^2 + d^2 + 2(ab + ac + ad + bc + bd + cd)$

For the latter part of the expression, we replace the 2ab and 2cd terms with our expressions above:

$a^2-a^2 + b^2-b^2 + c^2-c^2+d^2-d^2+(a+b)^2+(c+d)^2+2(ac+ad+bc+bd)$

We simply and factor the last term by grouping:

$(a+b)^2+(c+d)^2 +2(a+b)(c+d)$

$[(a+b)^2+(c+d)^2]^2$

This is our expression the square of our prime number, or $p^2 =(a+b+c+d)^2$ . So we can write:

$p=a+b+c+d = (a+b)^2 + (c+d)^2$

By the Fermat's Theorem, the number $a+b+c+d$ must also equal some square interger for positive intergers a,b,c,d. Notice, if $a+b+c+d$ is not an interger, then we have proof by contradiction. Therefore, a+b+c+d cannot be prime.

$\textbf{Q.E.D.}$

a+b+c+d is always going to be an even number. Of the 4 variables either none of them are odd, 2 of them are odd, or all four of them are odd. Which means the odds will cancel out when you add them up.

Since all four values are given to be integers Their sum must be an integer and products of two or more of any of these values must be integer.

You can rewrite two of the summands in terms of the other summands.

One can give the summands a common denominator without changing the expression's value. Dividing by this common denominator still gives an expression that is an integer. Therefore the expression can never be prime.

let a=a 1*a 2 and b=b 1*b 2, than c*d must contain all of components of a and b. If we calculate all possibilities, there are most be common dividable integer without 1 and a+b+c+d itself.

$a+b+c+d$ and $ab=cd$ .

Let's choose $b$ and $d$ to simplify our thought process. $b = n_1, d = n_2$ .

Using $a n_1 = c n_2$ we can quickly write our sum as $c \frac{n_2}{n_1} + n_2 + n_1 + c$ .

Simplify to $(n_1+n_2) ( \frac{c}{n_1} + 1)$

$n_1 \geq 1$ , $n_2 \geq 1$ , and $\frac{c}{n_1} + 1 > 1$ from our restriction that all numbers be positive integers.

So, all that's important here is that our number will always be divisible by $n_1 +n_2 \geq 2$ and therefore never prime.

A prime number can never be divisable by 2. If we have a+b+c+d = prime number, then if we divide by 2 like you would if ab=cd, then by definition the product cannot be a prime number.

For $a+b+c+d$ to be prime it has to be greater than 2, which means it is an odd prime.

For $a+b+c+d$ to be odd, one or three of the four numbers has to be even and the others has to be odd.

But than one side of the equation $ab=cd$ has to be even and the other side has to be odd, which is impossible, therefore it can not be prime.

Totally just guessed.

Assume the sum $a+b+c+d$ is odd. Then the residue sets modulo-2 of a,b,c,d can only be $\{0,0,0,1\}$ or $\{0,1,1,1\}$ . (i.e. Either one summand is odd and all the others even, or one summand is even and all the others odd.)

Given the 2 residue sets above, there is no way to combine the elements such that $ab (mod 2) = cd (mod 2)$ .

This contradicts our assumption that $a+b+c+d$ was odd.

Therefore $a+b+c+d$ must be even, and the only remaining prime value we need to consider is 2 . That cannot happen because we are given that $a,b,c,d$ are all positive integers so their sum must be $>= 4$ .

Therefore $a+b+c+d$ cannot be prime.

a=1, b=6, c=2, d=3 ab=cd 6=6 1+6+2+3=13 13 is a prime number............

So much for working.

First start with $ab=cd$

Add $bd$ to both sides:

$ab+bd=cd+bd$

Factor out the common factor:

$b(a+d)=d(c+b)$

Solve for $c+b$ :

$\frac{b}{d}(a+d)=c+b$

Now, when we try to evaluate $a+b+c+d$ , you see that you can plug in what we solved for:

$a+\frac{b}{d}(a+d)+d=(1+\frac{b}{d})(a+d)=(\frac{d+b}{d})(a+d)$

The only way this number can be prime, keeping in mind the conditions of the problem, is if $d$ completely divides any of the 2 factors. This can only be done if it equals the terms, which isn't the case since both terms have a $d$ in them, plus a positive integer. Thus, the sum is always composite given that $a$ , $b$ , $c$ , and $d$ are positive integers

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If a + b + c + d is prime, it must be odd, since 2 cannot be written as the sum of four positive integers. That means that either three of a, b, c and d are odd, or three of them are even. In either case, one of ab and cd will be odd and the other will be even. But ab = cd. So the sum cannot be prime.

Let's rewrite

• $a = 2^{a_1} * 3^{a_2} * 5^{a_3} * ...$ ;
• $b = 2^{b_1} * 3^{b_2} * 5^{b_3} * ...$ ;
• $c = 2^{c_1} * 3^{c_2} * 5^{c_3} * ...$ ;
• $d = 2^{d_1} * 3^{d_2} * 5^{d_3} * ...$ .

Then using the equation $ab = cd$ we can derive $a_1 + b_1 = c_1 +d_1$ , $a_2 + b_2 = c_2 + d_2$ etc, or $a_1 - c_1 = d_1 - b_1$ , $a_2 - c_2 = d_2 - b_2$ etc

Next (for the sake of simplicity only factors of "2" and "3" are shown),

$a + b + c + d =$ $= ( 2^{a_1} * 3^{a_2} ) + ( 2^{b_1} * 3^{b_2} ) + ( 2^{c_1} * 3^{c_2} ) + ( 2^{d_1} * 3^{d_2} ) =$ $= ( 2^{c_1} * 3^{c_2} ) ( 1 + 2^{a_1 - c_1 } * 3^{a_2 - c_2} ) + ( 2^{b_1} * 3^{b_2} ) ( 1 + 2^{d_1 - b_1 } * 3^{d_2 - b_2} )$

Using equation above:

$a + b + c + d = ( 1 + 2^{a_1 - c_1 } * 3^{a_2 - c_2} ) ( 2^{c_1} * 3^{c_2} + 2^{b_1} * 3^{b_2} )$

Thus the sum of these 4 numbers is always equal to a multiplication (of numbers greater than "1") and cannot be a Prime

If ab = cd, then both ab and cd must either be odd or even. In both cases, this requires the number of odd digits to equal 0, 2 or 4 (i.e. none of the numbers are odd, all the numbers are odd, or one variable from either side is odd). As an even number of odd digits always adds up to an even number, the sum of all the variables will always be divisible by 2, and therefore cannot be a prime number.

If a & b are both odd, then c & d both need to be odd. If at least one of a & b is even, then at least one of c & d is also even. So a+b+c+d is either a sum of 4 odd numbers (or) a sum of 2 odd and 2 even numbers (or) a sum of 4 even numbers. All of these possibilities end up with an even number for a+b+c+d. So a+b+c+d is even and hence not prime!

ab=cd so a=c and b=c - If a,b,c and d are odd numbers, the sum would be even, because odd numbers gathered even times will result an even sum. If a,b,c and d are even numbers, the sum would be even too. Because the sum would be even, the sum could be divided by 2. - And if we would have 3 odd numbers and one even, the sum would be odd too, but it will can be divided in an odd number. sorry for my bad english