# Introduction

How to prove that 1 = 2:

1 = 2
2 = 4 (*2)
-1 = 1 (-3)
1 = 1 (^2)


You can just multiply both sides by 0, but that's cheating. This is just bending the rules a little bit.

Write a program/function that, when given two integers, outputs a "proof" that the two are equal.

# Rules

• Your program's output must start with a = b, where a and b are the integers, and end with c = c, where c can be any number. Don't output the operations.
• Your output can only use integers.
• The only operations you can use are addition, multiplication, division, subtraction, and exponentiation, and you cannot multiply, divide or exponentiate by 0.
• Each line must be one of the above operations done to both sides of the equation in the previous line.

For example, you could go:

4 = 9
5 = 10 (+1)
1 = 2 (/5)
8 = 16 (*8)
-4 = 4 (-12)
16 = 16 (^2)


# Scoring

This is , shortest wins!

• Sandboxed. Going to sleep now, will try to clarify stuff later. Jun 1 at 11:46
• Alternatively phrased, this is equivalent to finding the steps from an input $(a, b)$ to $(-n, n)$ for some integer $n$, where the "steps" are the 5 provided operators Jun 1 at 11:55
• May we assume that a≠b? Jun 1 at 11:57
• From the sandbox: It's worth mentioning that for any a=b as input you can always do the steps a=b;2a=2b (*2); a-b=b-a (-(a+b));a^2-2ab+b2=a^2-2ab+b2 (^2). Sometimes you can do faster but there is not much reason to do anything more complex than (*2)->(-(a+b))->(^2). Jun 1 at 15:49
• Must I separate integers and an equal sign with a space? Jun 2 at 12:20

# JavaScript (Node.js), 57 54 bytes

a=>b=>a+=${b}${a+a}=${b+b}${a-=b}=${-a}${a*=a}=+a


Try it online!

## How?

There always exists a multiple of $$\0.5\$$ which, when subtracted from each number, creates a pair of the form $$\-a = a\$$. If $$\a-x=c\$$, then $$\b-(a-x+b)=-c\$$, so then if $$\a-x+b=x\$$, then $$\x=\frac{(a+b)}2\$$. This might not be an integer, so multiply by 2 beforehand.

-3 bytes thanks to Arnauld.

• 54 bytes Jun 1 at 12:18

# 05AB1E, 13 12 bytes

xIÂ-Dn)'=δý»


Try it online!

A port of my Jelly answer. I thought that 05AB1E's stack-based functionality would be shorter, but it isn't great at joining.

-1 byte thanks to ovs!

xIÂ-Dn)'=δý» - Full program. Takes [a, b] on the stack
x            - Push [2a, 2b];   Stack = [[a, b], [2a, 2b]]
I           - Push [a, b];     Stack = [[a, b], [2a, 2b], [a, b]]
Â          - Bifurcate;       Stack = [[a, b], [2a, 2b], [a, b], [b, a]]
-         - Subtract;        Stack = [[a, b], [2a, 2b], [a-b, b-a]]
D        - Duplicate;       Stack = [[a, b], [2a, 2b], [a-b, b-a], [a-b, b-a]]
n       - Square;          Stack = [[a, b], [2a, 2b], [a-b, b-a], [(a-b)², (b-a)²]]
)      - Wrap into array; Stack = [[[a, b], [2a, 2b], [a-b, b-a], [(a-b)², (b-a)²]]]
'=    - Push '='
δý  - Join each with '='
» - Join array by newlines

• A few 12-byters. '=δý uses double-vectorize (δ), the other ones use that » joins inner lists by spaces.
– ovs
Jun 1 at 14:52
• @ovs Nice! I think I'll go with the first one, δ seems a lot easier to understand than either # or :, especially in the context of » Jun 1 at 15:00

# R, 64 51 bytes

-7 bytes thanks to Dominic van Essen.

cat(sep=c("=","
"),x<-scan(),2*x,y<-2*x-sum(x),y^2)


Try it online!

The steps followed are:

a = b
2a = 2b (*2)
a-b = b-a (-(a+b))
(a-b)^2 = (b-a)^2 (^2)

• 55 bytes by changing step 3 to 'subtract both sides from a+b' instead of 'subtract a+b from both sides' and rearranging a bit... Jun 1 at 21:45
• @DominicvanEssen Thanks, but the way I read the rules, I think that would require 2 steps (first multiply by -1, then add a+b). Putting sep up front is a great idea! Jun 1 at 22:00
• This should be Ok though...? Jun 2 at 0:06
• @DominicvanEssen Very nice! Jun 2 at 13:11

# Jelly, 14 bytes

,Ḥ;_U,²$Ɗj€”=Y  Try it online! Uses ophact's observation, that for any $$\a, b\$$, it is sufficient to yield the 4 equations $$a = b \\ 2a = 2b \\ a-b = b-a \\ (a-b)^2 = (b-a)^2$$ ## How it works ,Ḥ;_U,²$Ɗj€”=Y - Main link. Takes [a, b] on the left
Ḥ             - Double, yielding [2a, 2b]
,              - Pair; [[a, b], [2a, 2b]]
U          -   Reverse; [b, a]
_           -   Subtract; [a-b, b-a]
$- Previous two links as a monad f([a-b, b-a]]): ² - Square; [(a-b)², (b-a)²] , - Pair; [[a-b, b-a], [(a-b)², (b-a)²]] ; - Concatenate; [[a, b], [2a, 2b], [[a-b, b-a], [(a-b)², (b-a)²]] j€”= - Join each with "=" Y - Join with newlines  # Python 2, 79 78 bytes for x in'input()','2*a,2*b','a-b>>1,b-a>>1','a*a,'*2:a,b=eval(x);print a,'=',b  Try it online! # Python 3, 70 bytes def f(a,b):x=a-b;return f'{a}={b}\n{2*a}={2*b}\n{x}={-x}\n{x*x}={x*x}'  Try it online! • I actually pointed out this method in the sandbox before ophact. But early bird gets the worm. Jun 1 at 15:54 • @WheatWizard I did not look at the sandbox before posting. – user100690 Jun 1 at 18:15 • With Python 3.8 you can get 64 bytes. Jun 1 at 22:09 # Python 2, 84 $$\\cdots\$$ 73 71 bytes Saved 5 bytes thanks to ovs!!! Saved 2 bytes thanks to dingledooper!!! k=a,b=input() d=a-b for a,b in k,[2*a,2*b],[d,-d],[d*d]*2:print a,'=',b  Try it online! • 73 bytes as a full program (or 75 if you keep it as function). – ovs Jun 1 at 20:23 • @ovs Nice one - thanks! :D Jun 1 at 20:56 • 71 bytes Jun 1 at 22:05 • 61 bytes Jun 2 at 0:16 • 54 bytes with Python 3.8's walrus operator. Jun 2 at 3:17 # C (clang), 65 bytes #define r;printf("%d=%d\n",a f(a,b){r,b)r+a,b+b)r-=b,-a)r*=a,a);}  Using clang instead of gcc saves a byte, as clang handles the undefined behavior in the second printf differently than gcc: printf("%d=%d\n",a-=b,-a);  The order of the arguments' evaluation is unspecified. In gcc, the -a is evaluated before a-=b; thus, the second argument is not affected by the subtraction and must be b-a to get the proper value. However, in clang, the a-=b is evaluated first so the second argument is affected by the subtraction so -a is the correct value. # C (gcc), 76 74 66 bytes -8 bytes thanks to tsh #define r;printf("%d=%d\n",a f(a,b){r,b)r+a,b+b)r-=b,b-a)r*=a,a);}  Try it online! • 66: #define r;printf("%d=%d\n",a f(a,b){r,b)r+a,b+b)r-=b,b-a)r*=a,a);} – tsh Jun 4 at 8:21 # Charcoal, 20 bytes ≔⁻⊗θΣθηＥ⟦θ⊗θηＸη²⟧⪫ι=  Try it online! Link is to verbose version of code. Takes input as a list [a, b]. Explanation: Based on @ophact's approach. ≔⁻⊗θΣθη  Vectorised subtract the sum of the list from its double, thus giving [a-b, b-a]. Ｅ⟦θ⊗θηＸη²⟧⪫ι=  Loop over the lists [a, b], 2[a, b], [a-b, b-a] and [a-b, b-a]², joining each list with = and implicitly printing them on separate lines. # C# (Visual C# Interactive Compiler), 50 bytes Same as the javascript but using C#'s "superior code golfing" interpolation a=>b=>a+@$"={b}
{a+a}={b+b}
{a-=b}={-a}
{a*=a}="+a


Try it online!

# Vyxalj, 12 bytes

:d?Ḃ-:²Wƛ\=j


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Why write your own original answer when you can just port 05ab1e amiright? :p

## Explained

:d?Ḃ-:²Wƛ\=j
:d           # [a, b], [2a, 2b]
?Ḃ-        # [a - b], [b - a]
:²      # that, but squared
Wƛ\=j # join each on "=" and then join that on newlines with the j flag


# Julia 1.0, 48 bytes

a\b="$a=$b
$(2a)=$(2b)
$(a-=b)=$(-a)
$(a*=a)=$a"


Try it online!

## alternative answer, 48 bytes too

output is a list of strings

a\b=join.([a=>b,2a=>2b,(a-=b)=>-a,a*a=>a*a],'=')


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# Python 3, 72 bytes

f=lambda a,b:f"{a}={b}\n{2*a}={2*b}\n{a-b}={b-a}\n{(a-b)**2}={(a-b)**2}"


Uses ophact's observation to answer in 4 steps.

Try it online!

• Nice! You can save two bytes by going a+f"= at the start and ="+(a-b)**2 at the end. Jun 8 at 4:24
• @Ausername That would be attempting to concatenate an integer and a string, which Python doesn't allow. Thanks anyway! Jun 8 at 10:50
• Sorry, I'm used to Javascript. Jun 8 at 11:02
• @Ausername Oh, ok. Javascript has a lot of strange type conversions (e.g. []+[] gives ""), so that makes sense. Jun 8 at 11:23

# GolfScript, 37 bytes

~:%;:$' = ':&%n$2*&%2*n$%-&%$-.n\2?.&\
`

Try it online!